I'm stuck on this Calculus problem about finding a parabola with the equation
$y=ax^2+bx+c$
I know the slope of $3$ at x=1 and slope of $-13$ at $x=-1$, and know that the parabola passes through the point $(1,1).$
Any help would be great.
[Math] Find equation of a parabola given slope of tangent line at two x values and a point on the curve
calculusderivativestangent line
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Best Answer
Let $f(x)=ax^{2}+bx+c$. We know the following:
\begin{cases} f'(1)=2a+b=3\\ f'(-1)=-2a+b=-13\\ f(1)=a+b+c=1 \end{cases}
This system has solution $a=4, b=-5$ and $c=2$. So the parabola you are looking for has equation $$y=4x^{2}-5x+2.$$