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

1 Answer
Jul 28, 2015

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...