Question

Write the equation in vertex form

y=3x^2 +8x+2

y= -2x^2+6x-2

Answer 1

Answer:

Part 1) $y=3(x+\frac{4}{3})^{2}-\frac{10}{3}$

Part 2) $y=-2(x-\frac{3}{2})^{2}+\frac{5}{2}$

Step-by-step explanation:

Part 1) we have

$y=3x^{2}+8x+2$

Factor the leading coefficient 3

$y=3(x^{2}+\frac{8}{3}x)+2$

Complete the square

$y=3(x^{2}+\frac{8}{3}x+\frac{64}{36})+2-\frac{64}{12}$

$y=3(x^{2}+\frac{8}{3}x+\frac{64}{36})-\frac{40}{12}$

Rewrite as perfect squares

$y=3(x+\frac{8}{6})^{2}-\frac{40}{12}$

simplify

$y=3(x+\frac{4}{3})^{2}-\frac{10}{3}$

The vertex is the point $(-\frac{4}{3},-\frac{10}{3})$

Part 2) we have

$y=-2x^{2}+6x-2$

Factor the leading coefficient -2

$y=-2(x^{2}-3x)-2$

Complete the square

$y=-2(x^{2}-3x+\frac{9}{4})-2+\frac{9}{2}$

$y=-2(x^{2}-3x+\frac{9}{4})+\frac{5}{2}$

Rewrite as perfect squares

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

The vertex is the point $(\frac{3}{2},\frac{5}{2})$