Question

What is the standard form? Given the vertex (-3,3) and the focus point (-3,2)

Answer 1

Answer:

$\large\boxed{y=-\dfrac{1}{4}x^2-\dfrac{3}{2}x+\dfrac{3}{4}}$

Step-by-step explanation:

The equation of a parabola in the vertex form:

$y=a(x-h)^2+k$

(h, k) - vertex

Then the focus is:

$\left(h,\ k+\dfrac{1}{4a}\right)$

We have:

$the\ vertex\ (-3,\ 3)\to h=-3,\ k=3\\\\the\ focus\ (-3,\ 2)\to h=-3,\ k+\dfrac{1}{4a}=2$

Substitute k = 3 and calculate a:

$3+\dfrac{1}{4a}=2\qquad\text{subtract 3 from both sides}\\\\\dfrac{1}{4a}=-1\\\\\dfrac{1}{4a}=\dfrac{-1}{1}\qquad\text{cross multiply}\\\\(-1)(4a)=(1)(1)\\\\-4a=1\qquad\text{divide both sides by (-4)}\\\\a=-\dfrac{1}{4}$

Substitute to the vertex form:

$y=-\dfrac{1}{4}(x-(-3))^2+3$

Convert ot the standard form

$y=ax^2+bx+c\\\\y=-\dfrac{1}{4}(x+3)^2+3\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\y=-\dfrac{1}{4}(x^2+2(x)(3)+3^2)+3\\\\y=-\dfrac{1}{4}(x^2+6x+9)+3\qquad\text{use the distributive property}\\\\y=\left(-\dfrac{1}{4}\right)(x^2)+\left(-\dfrac{1}{4}\right)(6x)+\left(-\dfrac{1}{4}\right)(9)+3\\\\y=-\dfrac{1}{4}x^2-\dfrac{6}{4}x-\dfrac{9}{4}+\dfrac{12}{4}\\\\y=-\dfrac{1}{4}x^2-\dfrac{3}{2}x+\dfrac{3}{4}$