Help Please

[Tinkermom]

The graph of the function f(x)=ax^2 + bx + c has vertex at (1,4) and passes through the point (-1,-8). Find a b c.

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ ax^2+bx+c, \ a \ parabola, \ given, \ find \ a,b, \ and \ c \ given \ vertex \ = \ (1,4) \ and \ f(x) \ passes$

$\displaystyle through \ point \ (-1,-8).$

$\displaystyle Hence \ since \ axis \ of \ symmetry, \ x \ =1, \ gives \ another \ point \ (3,-8)$

$\displaystyle f(1) \ = \ 4 \ = \ a+b+c$

$\displaystyle f(-1) \ = \ -8 \ = \ a-b+c$

$\displaystyle f(3) \ = \ -8 \ = \ 9a+3b+c$

$\displaystyle Solving \ the \ system \ gives \ a \ = \ -3, \ b \ = \ 6, \ and \ c \ = \ 1.$

$\displaystyle See \ graph.$

[attachment=0:1isidolz]fff.jpg[/attachment:1isidolz]

 

[soroban]

Hello, Tinkermom!

$\displaystyle \text{The graph of the function }\,f(x)\:=\:ax^2 + bx + c\:\text{ has its vertex at }(1,4)\,\text{and passes through the point }(-1,-8).$

$\displaystyle \text{Find }a, b \text{ and } c.$

$\displaystyle \text{The point }(1,4)\text{ is on the graph:}$
. . $\displaystyle f(1) = 4:\;\; a + b + c \:=\:4$ .[1]

\(\displaystyle \text{The point }(-1,-8)\text{ is on the graph:}}\)
. . $\displaystyle f(-1) = -8: \;\;a - b + c \:=\:-8$ .[2]

$\displaystyle \text{We know the vertix is at: }\:x \:=\:\frac{-b}{2a}$
. . $\displaystyle \text{So: }\;\frac{-b}{2a} \:=\:1 \quad\Rightarrow\quad 2a + b \:=\:0$ .[3]


$\displaystyle \text{We have a system of equations: }\;\begin{array}{ccccccc}a + b + c &=& 4 & [1] \\ a - b + c &=& -8 & [2] \\ 2a + b \qquad &=& 0 & [3] \end{array}$

$\displaystyle \text{Subtract [1] - [2]: }\;2b \:=\:12 \quad\Rightarrow\quad b \:=\:6$

$\displaystyle \text{Substitute into [3]: }\;2a + 6 \:=\:0 \quad\Rightarrow\quad a \:=\:-3$

\(\displaystyle \text{Substitute into [1]: }\;-3 + 6 + c \:=\:4 \quae\Rightarrow\quad c \:=\:1\)


$\displaystyle \text{Therefore: }\;f(x) \;=\;-3x^2 + 6x + 1$


Edit: Too slow . . . again!
.

 

[Tinkermom]

Thank you. This confirms, I was on the right track. Just needed help putting it together.

Thanks again.