# Thread: Tangent of a point

1. ## Tangent of a point

I attached the question number. I found the answer to A and B but I'm stuck on C.

For B I got y=x+7

but the tangent line for equation is 3x^2 -2x -4

Thank you.

2. ## Re: Tangent of a point

Oh, dear. $3x^2- 2x- 4$ is a parabola, NOT any line at all! What you mean is that the derivative of $x^3- x^2- 4x+ 4$ so gives the slope of the tangent line.

Given that, at x= a, the derivative is $3a^2- 2a- 4$ and the value is $a^3- a^2- 4a+ 4$. That means that the equation of the tangent line, at x= a, is $y= (3a^2- 2a- 4)(x- a)+ a^3- a^2- 4a+ 4$. Saying that line goes through (0, -8) mean that x= 0, y= -8 makes that a true equation: solve $(3a^2- 2a- 4)(0- a)+ a^3- a^2- 4a+ 4= -2a^3+ a^2+ 4= -8$.

3. ## Re: Tangent of a point

Originally Posted by minneola24
I attached the question number. I found the answer to A and B but I'm stuck on C.

For B I got y=x+7

but the tangent line for equation is 3x^2 -2x -4

Thank you.
for c,
remember that $(a,b)$ and $(0,-8)$ lie on the same line,i.e the tangent line. So the slope of the line is $\frac{8+b}{a}$.
Again you know that the slope of the $3x^2-2x-4$(which you got by differentiating ) and thus for $(a,b)$ the slope is $3a^2-2a-4$.
equate the slopes:
$\frac{8+b}{a}=3a^2-2a-4 \implies 8+b=3a^3-2a^2-4$ and $b=a^3-a^2-4a-4$ from the original equation:
so $8+a^3-a^2-4a-4=3a^3-2a^2-4a \implies 2a^3-a^2-12=0$ hence $a=2$ and $b=2^3-2^2-4(2)+4=0$