Question

Find the x intercepts of the parabola with vertex (4,75) and y intercept (0,27)

Answer 1

The equation for a parabola with vertex $(h, k)$ is
$y=a(x-h)^{2}+k$

We plug in:
$y=a(x-4)^{2}+75$

We find $a$ by plugging in the point $(0, 27)$:
$27=a(0-4)^{2}+75$
$\Rightarrow 27=a(-4)^{2}+75$
$\Rightarrow 27=16a+75$
$\Rightarrow 16a=-48$
$\Rightarrow a=-3$

So our equation is
$y=-3(x-4)^{2}+75$

To find x-intercepts, we set $y=0$:
$-3(x-4)^{2}+75=0$
$\Rightarrow -3(x-4)^{2}=-75$
$\Rightarrow (x-4)^{2}=25$
$\Rightarrow x-4=5, -5$
$\Rightarrow x=9, -1$

So the x-intercepts are 9 and -1.