How do you find the quadratic function with vertex (-5/2, 0) and point (-7/2, -16/3)?

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Answer 1 · 2017-05-04T18:43:36

Because you asked for a quadratic function, we shall discard the vertex form $x = a {(y - k)}^{2} + h$ $x=a(y-k)^2+h$ and use only:

$y = a {(x - h)}^{2} + k [1]$ $y = a(x-h)^2+k" [1]"$

where $(h, k)$ $(h,k)$ is the vertex.

Substitute $- \frac{5}{2}$ $-5/2$ for h and $0$ $0$ for k into equation [1]:

$y = a {(x - - \frac{5}{2})}^{2} + 0$ $y = a(x--5/2)^2+0$

Simplify:

$y = a {(x + \frac{5}{2})}^{2} [2]$ $y = a(x+5/2)^2" [2]"$

Substitute $- \frac{7}{2}$ $-7/2$ for x and $- \frac{16}{3}$ $-16/3$ for y into equation [2]:

$- \frac{16}{3} = a {(- \frac{7}{2} + \frac{5}{2})}^{2}$ $-16/3 = a(-7/2+5/2)^2$

Solve for a:

$- \frac{16}{3} = a {(- \frac{2}{2})}^{2} = a {(- 1)}^{2}$ $-16/3 = a(-2/2)^2 = a(-1)^2$

$a = - \frac{16}{3}$ $a = -16/3$

Substitute $- \frac{16}{3}$ $-16/3$ for a into equation [2]:

$y = - \frac{16}{3} {(x + \frac{5}{2})}^{2} [3]$ $y = -16/3(x+5/2)^2" [3]"$

$y = - \frac{16}{3} (x^{2} + 5 x + \frac{25}{4})$ $y = -16/3(x^2+5x+ 25/4)$

$y = - \frac{16}{3} x^{2} - \frac{80}{3} x - \frac{100}{3} [4]$ $y = -16/3x^2-80/3x-100/3" [4]"$

Equation [3] is the vertex form and equation [4] is the standard form.