Quadratic Equations: Standard and Vertex Form Practice AK

Find the vertex of each equation:

1. $x=\dfrac{-b}{2a}=\dfrac{6}{2(3)}=1$

$\eqalign{y&=3(1)^2-6(1)+5\\y&=3-6+5\\y&=2}$

vertex$=(1,2)$

2. $x=\dfrac{-b}{2a}=\dfrac{-8}{2(1)}=-4$

$\eqalign{y&=(-4)^2+8(-4)-3\\y&=16-32-3\\y&=-19}$

vertex$=(-4, -19)$

3. $x=\dfrac{-b}{2a}=-\dfrac{-1}{2(-2)}=\dfrac{-1}{4}$

$\eqalign{y&=-2(\dfrac{-1}{4})^2-\dfrac{-1}{4}+1\\-2(\dfrac{1}{16})-\dfrac{-1}{4}+1\\y&=-\dfrac{1}{8}+\dfrac{1}{4}+1\\y&=\dfrac{9}{8}}$

vertex$=(\dfrac{-1}{4},\dfrac{9}{8})$

4. $(h,k)=(7,-3)$

5. $(h,k)=(-2,1)$

6. $(h,k)=(6,-4)$ 

Rewrite each of the following equations in vertex and standard form. 

7. vertex (1,2), point (2, -5)

Vertex form:

$\eqalign{h=1, k=2\\y&=a(x-h)^2+k\\-5&=a(2-1)^2+2\\-5&=a+2\\a&=-7\\y&=-7(x-1)^2+2}$

Standard form:

$\eqalign{x&=\dfrac{-b}{2a}\\1&=\dfrac{-b}{2(-7)}\\1&=\dfrac{-b}{-14}\\b&=14\\y&=-7x^2+14x+c\\2&=-7(1)^2+14(1)+c\\2&=-7+14+c\\c&=-5\\y&=-7x^2+14x-5}$

8. vertex (3,6), y-intercept 2

Vertex form:

$\eqalign{h=3, k=6\\y&=a(x-h)^2+k\\2&=a(0-3)^2+6\\2&=9a+6\\a&=-\dfrac{4}{9}\\y&=-\dfrac{4}{9}(x-3)^2+6}$

Standard form:

$\eqalign{x&=\dfrac{-b}{2a}\\3&=\dfrac{-b}{2(-\dfrac{4}{9})}\\b&=\dfrac{8}{3}\\y&=-\dfrac{4}{9}x^2+\dfrac{8}{3}x+c\\\text{The y-intercept is 2, so }c=2\\y&=-\dfrac{4}{9}x^2+\dfrac{8}{3}x+2}$