Question

vertex of f(x)=4x^2+16x+7

Answer 1

Have you tried putting into vertex form?

Answer 2

\[f(x)=ax^2+bx+c \\ \text{ first step group first two terms together } \\ f(x)=(ax^2+bx)+c \\ \text{ Now I want to factor the coefficient of } x^2 \text{ from both of these grouped terms } \\ f(x)=(ax^2+\frac{a}{a}bx)+c \\ \text{ factor out the a from both of those grouped terms } \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ \text{ now here I'm just going to multiply } \frac{1}{a} \text{ and } b \\ f(x)=a(x^2+\frac{b}{a}x)+c \\ \text{ now I'm going to leave a little space } \\ f(x)=a(x^2+\frac{b}{a}x+?)+c -a \cdot ? \\ \text{ now you will notice this space is being multiplied by } a \\ \text{ but whatever you add in you subtract out and vice verse } a \cdot ? -a \cdot ? =0 \\ \text{ the ? needs to be something so we can write the thing in the ( ) as something } \\ \text{squared} \\ f(x)=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a \cdot (\frac{b}{2a})^2 \\ f(x)=a(x+\frac{b}{2a})^2+c-a \cdot \frac{b^2}{2^2 a^2} \\ f(x)=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\]