# How do you graph 16t^2+4t+7?

Jul 28, 2015

It's a parabola opening upward with vertex (low point) at $\left(t , y\right) = \left(- \frac{1}{8} , \frac{27}{4}\right)$.

#### Explanation:

You can complete the square to get the function in vertex form by first factoring a 16 out of the first two terms:

$16 {t}^{2} + 4 t + 7 = 16 \left({t}^{2} + \frac{1}{4} t\right) + 7$

Next, take the coefficient of $t$, which is $\frac{1}{4}$, divide it by 2 to get $\frac{1}{8}$, and square that to get $\frac{1}{64}$. Add this number inside the parentheses and then compensate by subtracting $16 \cdot \frac{1}{64} = \frac{1}{4}$ outside the parentheses:

$16 {t}^{2} + 4 t + 7 = 16 \left({t}^{2} + \frac{1}{4} t + \frac{1}{64}\right) + 7 - \frac{1}{4}$

The expression in the parentheses is now a perfect square:

$16 {t}^{2} + 4 t + 7 = 16 {\left(t + \frac{1}{8}\right)}^{2} + \frac{27}{4}$

This means that it's a parabola opening upward with vertex (low point) at $\left(t , y\right) = \left(- \frac{1}{8} , \frac{27}{4}\right)$.

If you let $f \left(t\right) = 16 {t}^{2} + 4 t + 7 = 16 {\left(t + \frac{1}{8}\right)}^{2} + \frac{27}{4}$. You can also plot other points to find, for example $f \left(0\right) = 7$, $f \left(- \frac{9}{8}\right) = f \left(\frac{7}{8}\right) = 16 + \frac{27}{4} = \frac{91}{4}$, etc...