what is $f(x)=2x^{2}+28x - 5$ written in vertex form?\n$f(x)=2(x + 7)^{2}-19$\n$f(x)=2(x +…

what is $f(x)=2x^{2}+28x - 5$ written in vertex form?\n$f(x)=2(x + 7)^{2}-19$\n$f(x)=2(x + 7)^{2}-103$\n$f(x)=2(x + 14)^{2}-14$\n$f(x)=2(x + 14)^{2}-98$

what is $f(x)=2x^{2}+28x - 5$ written in vertex form?\n$f(x)=2(x + 7)^{2}-19$\n$f(x)=2(x + 7)^{2}-103$\n$f(x)=2(x + 14)^{2}-14$\n$f(x)=2(x + 14)^{2}-98$

Answer

Explanation:

Step1: Factor out the coefficient of $x^{2}$

$f(x)=2(x^{2}+14x)-5$

Step2: Complete the square inside the parentheses

For $x^{2}+14x$, half of the coefficient of $x$ is $\frac{14}{2} = 7$, and $(x + 7)^{2}=x^{2}+14x + 49$. So we have $x^{2}+14x=(x + 7)^{2}-49$. Then $f(x)=2((x + 7)^{2}-49)-5$.

Step3: Expand and simplify

$f(x)=2(x + 7)^{2}-98 - 5=2(x + 7)^{2}-103$

Answer:

$f(x)=2(x + 7)^{2}-103$