• Nov 7th 2012, 04:21 AM
skweres1
The equation can be in any form (standard, vertex, or intercept)

A quadratic with zeros at -4 and -1 and y-intercept at -8
• Nov 7th 2012, 04:51 AM
HallsofIvy
A quadratic can be written in the form $\displaystyle y= ax^2+ bx+ c$. If "there is a zero at -4", then y= 0 when x= -4: $\displaystyle 0= a(-4)^2+ b(-4)+ c$. If "there is a zero at -1", then y= 0 when x= -1: $\displaystyle 0= a(-1)^2+ b(-1)+ c$. If "the y-intercept is at -8" then y= -8 when x= 0: $\displaystyle -8= a(0)^2+ b(0)+ c$.

That gives you three equations to solve for a, b, and c.
• Nov 7th 2012, 05:04 AM
Soroban
Hello, skweres1!

Quote:

The equation can be in any form (standard, vertex, or intercept)
. .
A quadratic with zeros at $\displaystyle \text{-}4$ and $\displaystyle \text{-}1$ and y-intercept at -8.

With zeros at $\displaystyle \text{-}1$ and $\displaystyle \text{-}4$,
. . the function has the form: .$\displaystyle f(x) \:=\:a(x+1)(x+4)$

We are told that $\displaystyle f(0) = \text{-}8\!:\;\;a(1)(4) \:=\:\text{-}8 \quad\Rightarrow\quad a \:=\:\text{-}2$

Therefore: .$\displaystyle f(x) \:=\:\text{-}2(x+1)(x+4) \:=\:\text{-}2x^2 - 10x - 8$