Question

Write y = 2x^2 + 8x + 3 in vertex form.

Answer 1

We'll use "completing the square" to rewrite y = 2x^2 + 8x + 3 in vertex form:

y = 2(x^2 + 4x              ) + 3

Halve the coefficient of x (that is, halve 4), obtaining 2, and then square 2.

Add this 2  and then subtract it as shown below:

y = 2(x^2 + 4x + 4 - 4             ) + 3

Next, rewrite x^2 + 4x + 4 as the square of a binomial:

y = 2( (x+2)^2 - 4             ) + 3

Multiply the -4 term by 2 and take the result outside the parentheses:

y = 2( (x+2)^2              ) + 3 - 8

Simplifying this result:

y = 2( (x+2)^2              )  - 5

Thus, the vertex is at x = -2 and y = -5:  (-2, -5).

Answer 2

answer:  $y = 2(x + 2)^{2} - 11$

explanation:

you gotta know the equation for vertex; $y = a(x - h)^{2} + k$. we can turn the quadratic equation into vertex using the complete square.

$y = 2(x^{2} + 4x) -3$  | first, the coefficient $x^{2}$ must be 1

then add/subtract ($\frac{1}{2}$ coffiecent of x term)^2 of $x^2 + 4x$

$y = 2(x^2 + 2 (2)x +4 - 4) -3$

$y = 2(x + 2)^2 - 8 -3$

$y = 2(x + 2)^2 -11$  | final answer

hope this helps! ❤from peachimin