Express $s = -5t^2 + 40t$ in the form of $a(t-b)^2 + c$, where $a$, $b$ and $c$ are the constants.

Question

$s= -5t^2+ 40t$.

Express $s$ in the form of $a(t-b)^2 + c$, where $a$, $b$ and $c$ are the constants.

$s = -5t(t-8)$. I have factorized it.

Answer 1

To transform the equation from standard form to vertex form, we must complete the square.

First, we write $$s = -5t^2 + 40t = -5(t^2 - 8t)$$ We need to transform $t^2 - 8t$ into a perfect square. Remember that $$(a + b)^2 = a^2 + 2ab + b^2$$ If we let $a = t$, then $2ab = 2bt = -8t \implies b = -4$. Thus, to create a perfect square inside the parentheses, we have to add $b^2 = (-4)^2 = 16$. However, $-5(t^2 - 8t + 16) \neq s$ since adding $16$ inside the parentheses adds $(-5)(16) = -80$ to $s$. To compensate, we must add $80$ outside the parentheses. Hence, \begin{align*} s & = -5t^2 + 40t\\ & = -5(t^2 - 8t)\\ & = -5(t^2 - 8t + 16) + 80\\ & = -5(t - 4)^2 + 80 \end{align*} where in the final step we have factored the expression $t^2 - 8t + 16$.

Answer 2

First, take out a factor of $-5$ for $$s=-5(t^2-8t)$$ and then complete the square to get $$s=-5((t-4)^2-16).$$ In completing the square I divide the insides of the bracket by $t$, then half the number and deduct the square of it from the whole thing. It's not a very mathematically rigorous explanation but for a layman and for quickness I hope you can see what I have done. If you want to be sure you can work out $(t-4)^2-16$ and and it will be $t^2-8t.$ The rest should be pretty straight forward, just multiply through by that $-5$ to get $$a=-5\qquad b=4\qquad c=80$$