Writing equations of PARABOLAS
Writing equations of PARABOLAS We can see on the graph that the roots of the quadratic are: x = 2 (since the graph cuts the x-axis at x = − 2); and x = 1 (since the graph cuts the x-axis at x = 1.) Now, we can write our function for the quadratic as follows (since if we solve the following for 0, we'll get our 2 intersection points): y = (x + 2)(x − 1) We can expand this to give: y = x2 + x − 2
This is a quadratic function which passes through the x-axis at the required points. But is this the correct answer? There are an infinite number of parabolas that passes through the points (−2, 0) and (1, 0).
Here are some of them (in green) in the graphs below;
Writing equations of PARABOLAS So how do we find the correct quadratic function for a given parabola (the one in blue in this case)? To find the unique quadratic function for our blue parabola, we need a third point on the curve. We can form 3 equations in 3 unknowns and solve them to get the required result via solving systems.
System of Equations On the original blue curve, we can see that it passes through the point (0, −3) on the y-axis. We'll use that as our 3rd known point in addition to (−2, 0) and (1, 0). Using our standard form of the quadratic, y = ax2 + bx + c, we substitute the known values for x and y.
Substituting (−2,0):
Substituting (1,0):
Substituting (0,−3):
0 = a(−2)2 + b(−2) + c
0 = a(1)2 + b(1) + c
−3 = a(0)2 + b(0) + c
= 4a − 2b + c
=a+b+c
=c So we get c = −3.
Substituting c = −3 in the first line gives: 4a − 2b = 3; and substituting into the second line gives: a + b = 3 Multiplying the last line by 2 gives: 2a + 2b = 6 Adding this to 4a − 2b = 3 gives: 6a = 9 This gives a = 1.5. Substituting a = 1.5 into a + b = 3, we get b = 1.5. So the correct quadratic function for the blue graph is y = 1.5x2 + 1.5x − 3
Try This Find the equation of the parabola that goes through (−2, 4), (1, 1) and (3, 9).
Writing equations of PARABOLAS Another way of going about this is to observe the vertex of the parabola. We can write any quadratic equation for a parabola in “vertex form” as follows: y = a(x − h)2 + k
Vertex Form If the vertex of the parabola can be observed, then we get the required result via the vertex form.
This parabola has its vertex at (1, 0). If we use y = a(x − h)2 + k, we can see from the graph that h = 1 and k = 0. This gives us y = a(x − 1)2. So what is the value of “a”?
But as in the previous case, we have an infinite number of parabolas passing through (1, 0). Here are some of them on the right (sketched in green): In this example, the blue curve passes through (0, 1) on the y-axis, so we can simply substitute x = 0, y = 1 into y = a(x − 1)2 as follows: 1 = a(− 1)2 This gives us a = 1.
So our quadratic function for this example is y = (x − 1)2 = x2 − 2x + 1
Writing equations of PARABOLAS Here is another example
We can see the vertex is at (-2, 1) and the yintercept is at (0, 2). We just substitute as before into the vertex form of our quadratic function. We have (h, k) = (-2, 1) and at x = 0, y = 2. So y = a(x − h)2 + k Becomes 2 = a(0 − (−2))2 + 1 2 = 4a +1 a = 0.25 So our quadratic function y = 0.25(x − (−2))2 + 1 = 0.25(x + 2)2 + 1 Try This Find the equation of the parabola with vertex (−2, 4), and goes through (3, 9).
EXERCISES a) Find an equation of a vertical parabola through the given points and/or vertex. b) Find an equation of a horizontal parabola through the given points and/or vertex. 1) 2) 3) 4)
(2,5), (6,5), and (4,9). (2,3), (6, 5), and (2, 9). (1,3), and vertex (6, 5). (2, 4), and vertex (2, 9).
For Vertical Parabola
For Horizontal Parabola
Standard Form: y ax2 bx c
Standard Form: x ay 2 by c
Vertex Form: y a x h k
Vertex Form: x a y k h
2
2
This is a quadratic function which passes through the x-axis at the required points. But is this the correct answer? There are an infinite number of parabolas that passes through the points (−2, 0) and (1, 0).
Here are some of them (in green) in the graphs below;
Writing equations of PARABOLAS So how do we find the correct quadratic function for a given parabola (the one in blue in this case)? To find the unique quadratic function for our blue parabola, we need a third point on the curve. We can form 3 equations in 3 unknowns and solve them to get the required result via solving systems.
System of Equations On the original blue curve, we can see that it passes through the point (0, −3) on the y-axis. We'll use that as our 3rd known point in addition to (−2, 0) and (1, 0). Using our standard form of the quadratic, y = ax2 + bx + c, we substitute the known values for x and y.
Substituting (−2,0):
Substituting (1,0):
Substituting (0,−3):
0 = a(−2)2 + b(−2) + c
0 = a(1)2 + b(1) + c
−3 = a(0)2 + b(0) + c
= 4a − 2b + c
=a+b+c
=c So we get c = −3.
Substituting c = −3 in the first line gives: 4a − 2b = 3; and substituting into the second line gives: a + b = 3 Multiplying the last line by 2 gives: 2a + 2b = 6 Adding this to 4a − 2b = 3 gives: 6a = 9 This gives a = 1.5. Substituting a = 1.5 into a + b = 3, we get b = 1.5. So the correct quadratic function for the blue graph is y = 1.5x2 + 1.5x − 3
Try This Find the equation of the parabola that goes through (−2, 4), (1, 1) and (3, 9).
Writing equations of PARABOLAS Another way of going about this is to observe the vertex of the parabola. We can write any quadratic equation for a parabola in “vertex form” as follows: y = a(x − h)2 + k
Vertex Form If the vertex of the parabola can be observed, then we get the required result via the vertex form.
This parabola has its vertex at (1, 0). If we use y = a(x − h)2 + k, we can see from the graph that h = 1 and k = 0. This gives us y = a(x − 1)2. So what is the value of “a”?
But as in the previous case, we have an infinite number of parabolas passing through (1, 0). Here are some of them on the right (sketched in green): In this example, the blue curve passes through (0, 1) on the y-axis, so we can simply substitute x = 0, y = 1 into y = a(x − 1)2 as follows: 1 = a(− 1)2 This gives us a = 1.
So our quadratic function for this example is y = (x − 1)2 = x2 − 2x + 1
Writing equations of PARABOLAS Here is another example
We can see the vertex is at (-2, 1) and the yintercept is at (0, 2). We just substitute as before into the vertex form of our quadratic function. We have (h, k) = (-2, 1) and at x = 0, y = 2. So y = a(x − h)2 + k Becomes 2 = a(0 − (−2))2 + 1 2 = 4a +1 a = 0.25 So our quadratic function y = 0.25(x − (−2))2 + 1 = 0.25(x + 2)2 + 1 Try This Find the equation of the parabola with vertex (−2, 4), and goes through (3, 9).
EXERCISES a) Find an equation of a vertical parabola through the given points and/or vertex. b) Find an equation of a horizontal parabola through the given points and/or vertex. 1) 2) 3) 4)
(2,5), (6,5), and (4,9). (2,3), (6, 5), and (2, 9). (1,3), and vertex (6, 5). (2, 4), and vertex (2, 9).
For Vertical Parabola
For Horizontal Parabola
Standard Form: y ax2 bx c
Standard Form: x ay 2 by c
Vertex Form: y a x h k
Vertex Form: x a y k h
2
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