WEBVTT
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Let's talk about quadratic functions.
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One thing we'd like to be able to do is draw the graph of a quadratic function.
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Let's look at some features that will help us to draw the graph.
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First of all, this is a quadratic function
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since we see the x squared
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and no other higher powers of x.
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We know that the shape of the graph is going to be a parabola,
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and we also know the domain is going to be the set of all real numbers.
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Perhaps the most important feature for the parabola is its vertex, so let's find that.
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We can use this formula: negative b over 2a, f of negative b over 2a.
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Of course we need to know what are a and b.
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Those are going to come from our quadratic function.
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The letter a is the coefficient of x squared,
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and b is the coefficient of x.
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Let's find negative b over 2a first.
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Here, b is negative 2,
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so this will be a negative, negative 2
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over 2 times a. Here, a is 1.
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If we simplify this, this is positive 2 over 2 or 1.
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The next thing I need to find then is the y-coordinate of the vertex.
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I'm going to do that by using 1, the x value that we just found,
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and find f of 1.
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I substitute this for the x in the original problem.
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1 squared minus 2 times 1 minus 3 equals negative 4.
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My vertex is the point (1, -4).
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Let's go ahead and put that on our graph.
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A second feature we should discuss
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is which way the parabola is oriented.
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Since the coefficient of x squared is positive,
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we know that this parabola should open up.
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We can have an idea of the shape of the parabola here,
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but we need some more information to make it accurate.
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Another important feature of any graph would be intercepts.
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Let's see if we have any.
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We'll start with the y-intercept.
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The y-intercept is the y-coordinate of any point
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where the graph touches or crosses the y-axis.
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To find it, we let x equal 0.
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We need to find here f of 0, which is negative 3.
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There's my y-intercept at (0, -3).
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Let's plot the point. Let's look for x-intercepts.
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The x-intercept is the x-coordinate of any point
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where the graph crosses the x-axis.
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To find them, let y equal 0.
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Solve for x. Substitute 0 for y or f of x.
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This one we can factor.
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We want x minus 3 times x plus 1.
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This gives me two x-intercepts: 3 and negative 1.
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I can plot these points on my graph.
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Now remember the graph is the shape of a parabola,
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so now I'm going to connect these
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with a smooth, parabola-shaped curve.
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This is my function f. Now that we have the graph,
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there are many other features that we can discover
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about this function.
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Now that we've graphed this quadratic function,
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let's look at other features.
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First of all, we notice that our graph has some symmetry.
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We can draw in the line of symmetry for our graph.
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It is a vertical line passing through the vertex.
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The equation of a vertical line is always
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x equals the x-intercept. This is x equals 1.
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We can also find the range.
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The range is values that y is allowed to have.
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If we look at this, we see that the lowest y value occurs at the vertex,
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and the y value there is negative 4.
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Anything larger than negative 4 is also allowed.
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There is one final question we want to answer.
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On what intervals is the function increasing or decreasing?
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We want to look from left to right
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as we answer this question.
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Starting at the vertex, as the x values get larger,
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notice that the function values are rising.
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They're increasing. f is increasing starting at the vertex,
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which has an x-coordinate of 1,
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and as x gets larger without bound.
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On this portion of the graph,
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we see that the y values are getting smaller
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as x gets closer to the vertex.
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The x-coordinate of the vertex is 1,
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so f is decreasing on negative infinity to 1.
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Let's look at another example.
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Here's another quadratic function. Let's look at its features.
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The first thing I notice here
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is that the coefficient of x squared is a negative number.
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That tells me this parabola is going to open down.
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Another thing I should write down right away
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is the domain of this quadratic function.
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The domain is all real numbers. Let's find the vertex.
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We're going to proceed by finding the x-coordinate of the vertex:
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negative b over 2a.
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Remember that b is the coefficient of x,
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and a is the coefficient of the x squared term.
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If we simplify this, 6 over negative 6 is negative 1.
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To find the y-coordinate of the vertex,
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I need to find f of negative 1.
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If we work this out, this is negative 3 plus 6 minus 2, which is 1.
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Now I can go back and write the coordinates of the vertex,
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which are (-1, 1).
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Let's plot that on our graph.
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Now we've already said that this parabola will open down,
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so let's picture the parabola on our graph.
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It opens down, so I can see it's going to eventually cross the y-axis,
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and it's going to cross the x-axis twice.
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I know I need to find the intercepts.
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To find the y-intercept, we let x equal 0.
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We want to find f of 0. That's going to be 0 minus 0 minus 2.
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That's negative 2.
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Let's put our y-intercept on our graph: (0, -2).
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Now let's look for our x-intercepts.
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To find an x-intercept, you let y equal 0.
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We're going to substitute 0 for y or f of x,
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and we're going to try to solve this equation.
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Now I'd like to factor this, but it doesn't factor easily.
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Let's use the quadratic formula:
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x must be equal to negative b, which will be a negative, negative 6,
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plus or minus the square root of
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b squared, which is negative 6 squared, minus 4
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times a, which is negative 3, times c, which is negative 2,
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and all of this is over 2 times a,
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which we said is negative 3. Okay, now simplify this.
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This is positive 6 plus or minus the square root of
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negative 6 squared, which is 36,
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and here I have negative 4 times negative 3 times negative 2.
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That's three negative numbers, so the result is negative.
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4 times 3 is 12 times 2 is 24.
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2 times negative 3 in the denominator is negative 6.
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This is 6 plus or minus the square root of 12 over negative 6.
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We want the x-intercepts in order to draw the graph.
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We need to know approximately where to put them on the graph,
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so let's use our calculator and approximate the x-intercepts.
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The first one is open parentheses 6 plus
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the square root of 12.
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Close your parentheses, and then divide that by negative 6.
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This is approximately negative 1.58. Now let's find the other one.
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Again, open parentheses 6 minus the square root of 12.
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Close your parentheses, and then divide this by negative 6.
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This one is approximately negative 0.42.
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We're not going to be able to locate them precisely on our graph,
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but we can get them in the right area. Negative 1.58 is between negative 1 and negative 2,
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and negative 0.42 would be approximately here.
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Remember that the shape is a parabola that opens downward.
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You want to make sure this is a nice smooth curve.
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This is the graph of f. Now let's look at some other features of this graph.
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The axis of symmetry is a vertical line passing through the vertex of the parabola.
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Let's sketch that on our graph first.
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Remember the equation of this vertical line is x equals the x-value of the vertex.
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The axis of symmetry is the vertical line x equals negative 1.
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Let's find the range. The range is the set of values y is allowed to have.
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We see on this function that it has y-values all the way up to and including positive 1.
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We'll write the interval from negative infinity to positive 1
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and close it with a bracket. The last question we like to answer
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is the intervals on which the function f is increasing
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and those on which it's decreasing.
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Starting on the lefthand side,
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notice that the function values are rising
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until you get to the vertex.
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We say that f is increasing on the interval
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from negative infinity to the x-coordinate of the vertex.
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That would be negative 1.
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At the vertex, the graph turns,
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and the function values begin to decrease.
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We say that f is decreasing from negative 1 to positive infinity.
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Now let's look at some other examples of quadratic functions.
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This next example gives us the graph of a quadratic function
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and asks us to write the function algebraically.
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It's worth noting here that, up until this point,
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we've always looked at quadratic functions
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in the form f of x equals a times x squared plus b times x plus c where a is not 0.
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There is another form of the quadratic function,
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and that's called the standard form.
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In the standard form, the nice thing is that we can see the vertex
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of the quadratic function: h and k.
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Then we have this one other value a that we're going to have to find.
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In the given graph, we can pick out the vertex.
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The vertex has coordinates (2, 1).
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We can immediately substitute those values for h and k.
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The letter h is 2, and k is 1.
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As you can see, we're almost there.
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We just have that one missing value, a, that we need to find.
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In order to find it, we're going to use the one other given point on our graph.
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The function passes through the point (1, 3).
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This is an x and a y value that correspond to one another
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and make this a true statement.
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Remember that f of x is just another name for y,
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so we're going to replace f of x with 3.
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Then we want to replace that x with 1,
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the coordinate of that given point. Now let's solve for a.
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We need to subtract 1 from both sides
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and get 3 minus 1, which is 2.
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Negative 1 squared is 1, so we find out that a is 2.
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We can make this substitution here for a,
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and we find that our quadratic function is 2
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times x minus 2 squared plus 1.
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This is in standard form, and, sometimes, that's enough.
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Often, you're going to be asked to go ahead and go a little farther
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and write this in the other form:
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f of x equals a times x squared plus b times x plus c where a is not 0.
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In order to do that, we just need to multiply this out
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and get rid of the parentheses. Remember the order of operations.
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You have to do the squaring before you can distribute the 2.
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This is x minus 2 times itself.
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That's x squared minus 4x plus 4 plus 1.
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Now we can distribute.
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The last step is to combine those last two numbers.
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Here we have two versions of the equation for this quadratic function:
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the standard form and the a times x squared plus b times x plus c form.
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This example says, without graphing,
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to determine whether the quadratic function has a maximum or minimum value
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and then to tell what that value is.
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We don't want to draw the graph,
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but we probably should think about it a little bit.
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This quadratic function has a positive coefficient of x squared.
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That means that the graph is a parabola that opens up.
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I'm going to think about that parabola.
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The vertex is the lowest point on the graph,
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so I know this one must have a minimum value.
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I also know that the minimum value is
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the y-coordinate of the vertex, and that's what I need to find. Okay.
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We know how to find the coordinates of the vertex.
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Let's do it. To find the x-coordinate, we want to find
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negative b over 2a. Here, b is 12,
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so we get negative 12 over 2 times 2. Negative 12 over 4 is negative 3.
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The next step is to take this and find f of negative 3.
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2 times negative 3 squared plus 12 times negative 3 minus 1.
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2 times 9 is 18, minus 36, minus 1.
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This works out to be negative 19.
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That's the y-coordinate of the vertex,
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and for this quadratic function it is the minimum value.
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Now we have one more application.
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The price p and quantity x sold
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of a certain product obey the demand equation:
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p equals negative 1/6 x plus 80 for x between 0 and 480.
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The first thing we want to do is express the revenue as a function of x.
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We need to remember that revenue is price times quantity.
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If we're writing revenue as a function of x,
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price times quantity becomes p times x.
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Price is negative 1/6 x plus 80.
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Multiply times the quantity x. We're going to distribute the x.
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The next thing we want to do is find the quantity x that maximizes our revenue,
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what that maximum revenue is, and the price that should be charged
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to get the maximum revenue.
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We look at our revenue function. This is a quadratic function.
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The coefficient of x squared is negative,
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so we know the graph is going to be a parabola
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that opens down. Let's think about that shape.
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This function does in fact have a maximum value.
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It's going to occur at the vertex.
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We need to find the x-coordinate of the vertex
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to answer the first part.
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The maximum revenue is the y-coordinate of the vertex.
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Then we'll worry about finding the price.
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To find the x-coordinate of the vertex,
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we want to do negative b over 2a.
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That's going to be negative 80
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over 2 times negative 1/6.
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This is going to end up being a positive number,
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and it's 80 divided by 1/3. That's actually 240.
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The next thing: Find the revenue when x is 240.
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R of 240 is negative 1/6 times 240 squared plus 80
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times 240. We need to work this out on our calculator.
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Now with our calculator, let's work this out.
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For negative 1/6, I'm going to use the fraction key on the calculator.
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Times 240 squared plus 80 times 240.
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This tells me the maximum revenue is 9,600 dollars.
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There's one more part to answer.
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We want to know the price that should be charged
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to maximize the revenue.
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We're going to go back to our price demand equation.
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We know that when x is 240, we get the maximum revenue,
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so we're going to find the price when x is 240.
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Again, we'll use our calculator,
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or maybe we don't have to. 1/6 of 240 is 40, so we have negative 40 plus 80
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is 40. The company should charge 40 dollars for each product.
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This is it for quadratic functions.
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Now you need to review these problems
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and continue practicing to gain skill and mastery.