Determine a quadratic function f(x)=ax^2+bx+c

[Guest]

Hi,
I need help with this question

Determine a quadratic function f(x)=ax^2+bx+c whose graph passes through the point (2,19) and that has a horizontal tangent at (-1,-8)

I have no idea how to do it.

thanks for the help

 

[pka]

\(\displaystyle \begin{array}{rcl}
f(x) & = & ax^2 + bx + c\quad \Rightarrow \quad f'(x) = 2ax + b \\
19 & = & a\left( 2 \right)^2 + b\left( 2 \right) + c \\
- 8 & = & a\left( { - 1} \right)^2 + b\left( { - 1} \right) + c \\
0 & = & 2a\left( { - 1} \right) + b \\
\end{array}\)

 

[soroban]

Hello, bittersweet!

Determine a quadratic function \(\displaystyle f(x)\:=\:ax^2\,+\,bx\,+\,c\)
whose graph passes through the point (2,19) and that has a horizontal tangent at (-1,-8)

We need to determine the values of \(\displaystyle a,\,b,\,c\) . . . three variables.


We are given two points: \(\displaystyle (2,19)\) and \(\displaystyle (-1,8)\).

These give us: \(\displaystyle \,a\cdot2^2\,+\,b\cdot2\,+\,c\:=\:19\;\;\Rightarrow\;\;4a\,+\,2b\,+\,c\:=\:19\;\) [1]

\(\displaystyle \;\;\;\)and: \(\displaystyle \,a\cdot(-1)^2\,+\,b(-1)\,+\,c\:=\:8\;\;\Rightarrow\;\;a\,-\,b\,+\,c\:=\:8\;\) [2]


We seem to have only two equations . . . but there is a third equation.

We're told that there is a horizontal tangent at (-1,8).
\(\displaystyle \;\;\)This means that the derivative is zero at \(\displaystyle x\,=\,-1.\)

Since \(\displaystyle f'(x)\:=\;2ax\,+\,b\), we have: \(\displaystyle \,2a(-1)\,+\,b\:=\:0\;\;\Rightarrow\;\;-2a\,+\,b\:=\:0\.\) [3]


Now, we can solve the system of three equations . . .

\(\displaystyle \;\;\)I got: \(\displaystyle \,a\,=\,\frac{11}{9},\;y\,=\,\frac{22}{9},\;c\,=\,\frac{83}{9}\)
.