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Help Please
- Thread starter Tinkermom
- Start date
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
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- 1,577
\(\displaystyle f(x) \ = \ ax^2+bx+c, \ a \ parabola, \ given, \ find \ a,b, \ and \ c \ given \ vertex \ = \ (1,4) \ and \ f(x) \ passes\)
\(\displaystyle through \ point \ (-1,-8).\)
\(\displaystyle Hence \ since \ axis \ of \ symmetry, \ x \ =1, \ gives \ another \ point \ (3,-8)\)
\(\displaystyle f(1) \ = \ 4 \ = \ a+b+c\)
\(\displaystyle f(-1) \ = \ -8 \ = \ a-b+c\)
\(\displaystyle f(3) \ = \ -8 \ = \ 9a+3b+c\)
\(\displaystyle Solving \ the \ system \ gives \ a \ = \ -3, \ b \ = \ 6, \ and \ c \ = \ 1.\)
\(\displaystyle See \ graph.\)
[attachment=0:1isidolz]fff.jpg[/attachment:1isidolz]
\(\displaystyle through \ point \ (-1,-8).\)
\(\displaystyle Hence \ since \ axis \ of \ symmetry, \ x \ =1, \ gives \ another \ point \ (3,-8)\)
\(\displaystyle f(1) \ = \ 4 \ = \ a+b+c\)
\(\displaystyle f(-1) \ = \ -8 \ = \ a-b+c\)
\(\displaystyle f(3) \ = \ -8 \ = \ 9a+3b+c\)
\(\displaystyle Solving \ the \ system \ gives \ a \ = \ -3, \ b \ = \ 6, \ and \ c \ = \ 1.\)
\(\displaystyle See \ graph.\)
[attachment=0:1isidolz]fff.jpg[/attachment:1isidolz]
Hello, Tinkermom!
\(\displaystyle \text{The point }(1,4)\text{ is on the graph:}\)
. . \(\displaystyle f(1) = 4:\;\; a + b + c \:=\:4\) .[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!
.
\(\displaystyle \text{The graph of the function }\,f(x)\:=\:ax^2 + bx + c\:\text{ has its vertex at }(1,4)\,\text{and passes through the point }(-1,-8).\)
\(\displaystyle \text{Find }a, b \text{ and } c.\)
\(\displaystyle \text{The point }(1,4)\text{ is on the graph:}\)
. . \(\displaystyle f(1) = 4:\;\; a + b + c \:=\:4\) .[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!
.