Question # 1
Answer:
Therefore, the vertex form is:
[tex]f(x)=15(x^2+2)^2-79[/tex]
And the vertex of the function is: (-2, -79)
Step-by-step explanation:
Given the function
[tex]f\left(x\right)=15x^2+60x-19[/tex]
Factoring 15
[tex]f\left(x\right)=15\left(x^2+4x\right)-19[/tex]
Adding and subtracting the square of half the coefficient of [tex]x[/tex].
[tex]f\left(x\right)=15\left(x^2+4x+4\right)-19-4\times 15[/tex]
[tex]f\left(x\right)=15\left(x^2+4x+4\right)-19-60[/tex]
[tex]f\left(x\right)=15\left(x^2+2\right)^2-79[/tex]
Therefore, the vertex form is:
[tex]f(x)=15(x^2+2)^2-79[/tex]
And the vertex of the function is: (-2, -79)
Question # 2
Answer:
- [tex]f(x)=8(x-\frac{1}{4})^2 +\frac{21}{2}[/tex] is the vertex form of [tex]f(x)=8x^2-4x+11[/tex].
Step-by-step explanation:
Given the function
[tex]f(x)=8x^2-4x+11[/tex]
Factoring 8 from the first two terms
[tex]f\left(x\right)=8\left(x^2-\frac{1}{2}\:x\right)+11[/tex]
Next adding and subtracting the square of half the coefficient of the linear term
[tex]f\left(x\right)=8\left(x^2-\frac{1}{2}\:x+\frac{1}{16}\:\right)+11-8\times \frac{1}{16}[/tex]
[tex]f\left(x\right)=8\left(x^2-\frac{1}{2}\:x+\frac{1}{16}\:\right)+11-\:\frac{1}{2}[/tex]
Factoring the perfect square trinomial
[tex]f\left(x\right)=8\left(x-\frac{1}{4}\right)^2\:+\frac{21}{2}[/tex]
Therefore,
[tex]f(x)=8(x-\frac{1}{4})^2 +\frac{21}{2}[/tex] is the vertex form of [tex]f(x)=8x^2-4x+11[/tex].
