Question

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

Answer 1

It's a parabola opening upward with vertex (low point) at `(t,y) = (-1/8,27/4)`.

Explanation:

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

`16t^2+4t+7=16(t^2+1/4 t)+7`

Next, take the coefficient of `t`, which is `1/4`, divide it by 2 to get `1/8`, and square that to get `1/64`. Add this number inside the parentheses and then compensate by subtracting `16*1/64=1/4` outside the parentheses:

`16t^2+4t+7=16(t^2+1/4 t+1/64)+7-1/4`

The expression in the parentheses is now a perfect square:

`16t^2+4t+7=16(t+1/8)^2+27/4`

This means that it's a parabola opening upward with vertex (low point) at `(t,y) = (-1/8,27/4)`.

If you let `f(t)=16t^2+4t+7=16(t+1/8)^2+27/4`. You can also plot other points to find, for example `f(0)=7`, `f(-9/8)=f(7/8) = 16+27/4=91/4`, etc...