Answer
The equation of the parabola is $$y=f(x)=3x^2-2x+7$$
Work Step by Step
$$y=f(x)=ax^2+bx+c$$
- The derivative of $f(x)$ would be $$f'(x)=2ax+b$$
1) The slope of the tangent line of $f(x)$ at $A(x,y)$ would equal $f'(x)$
According to the question:
- $f(x)$ has slope 4 at $x=1$. That means $f'(1)=4$. So, $$2a\times1+b=4$$ $$2a+b=4\hspace{1cm}(1)$$
- $f(x)$ has slope -8 at $x=-1$. That means $f'(-1)=-8$. So, $$2a\times(-1)+b=-8$$ $$-2a+b=-8$$ $$2a-b=8\hspace{1cm}(2)$$
Summing (1) and (2), we have $$4a=12$$ $$a=3$$
Therefore, $b=2a-8=2\times3-8=-2$
So, $y=f(x)=3x^2-2x+c$
2) The parabola passes through the point $(2,15)$. Therefore, $$3\times2^2-2\times2+c=15$$ $$8+c=15$$ $$c=7$$
Therefore, the equation of the parabola is $$y=f(x)=3x^2-2x+7$$
