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.

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- Feb 6th 2013, 08:51 AMminneola24Tangent 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. - Feb 6th 2013, 09:51 AMHallsofIvyRe: Tangent of a point
Oh, dear. is a parabola, NOT any line at all! What you mean is that the

**derivative**of so gives the slope of the tangent line.

Given that, at x= a, the derivative is and the value is . That means that the equation of the tangent line, at x= a, is . Saying that line goes through (0, -8) mean that x= 0, y= -8 makes that a true equation: solve . - Feb 6th 2013, 09:55 AMearthboyRe: Tangent of a point
for c,

remember that and lie on the same line,i.e the tangent line. So the slope of the line is .

Again you know that the slope of the (which you got by differentiating ) and thus for the slope is .

equate the slopes:

and from the original equation:

so hence and