Question

Please help me! No one has answered me all night and I'm in quite of a bind.

Answer 1

                         Question # 1

Answer:

Therefore, the vertex form is:

$f(x)=15(x^2+2)^2-79$

And the vertex of the function is: (-2, -79)

Step-by-step explanation:

Given the function

$f\left(x\right)=15x^2+60x-19$

Factoring 15

$f\left(x\right)=15\left(x^2+4x\right)-19$

Adding and subtracting the square of half the coefficient of $x$.

$f\left(x\right)=15\left(x^2+4x+4\right)-19-4\times 15$

$f\left(x\right)=15\left(x^2+4x+4\right)-19-60$

$f\left(x\right)=15\left(x^2+2\right)^2-79$

Therefore, the vertex form is:

$f(x)=15(x^2+2)^2-79$

And the vertex of the function is: (-2, -79)

                       Question # 2

Answer:

• $f(x)=8(x-\frac{1}{4})^2 +\frac{21}{2}$  is the vertex form of $f(x)=8x^2-4x+11$.

Step-by-step explanation:

Given the function

$f(x)=8x^2-4x+11$

Factoring 8 from the first two terms

$f\left(x\right)=8\left(x^2-\frac{1}{2}\:x\right)+11$

Next adding and subtracting the square of half the coefficient of the linear term

$f\left(x\right)=8\left(x^2-\frac{1}{2}\:x+\frac{1}{16}\:\right)+11-8\times \frac{1}{16}$

$f\left(x\right)=8\left(x^2-\frac{1}{2}\:x+\frac{1}{16}\:\right)+11-\:\frac{1}{2}$

Factoring the perfect square trinomial

$f\left(x\right)=8\left(x-\frac{1}{4}\right)^2\:+\frac{21}{2}$

Therefore,

$f(x)=8(x-\frac{1}{4})^2 +\frac{21}{2}$  is the vertex form of $f(x)=8x^2-4x+11$.