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I got this question wrong on my homework last week, I was hoping you could show me how to do it.

Given: f(x)=3x^2+5x-2

Given that equation, I am asked to find:

Part A) the slope of f(x) at x=2

Part B) an equation for the tangent to f(x) at x=2

Part C) an equation for the normal line to f(x) at x=2

I did 9 problems exactly like this and got all of the parts right on all of them, but completely bombed this one. Was hoping you could show me what the right answers are.Thank you for your time.

To find the slope at any point on a curve, find the first derivative and then evaluate it at the specific x value

f'(x) = 6x + 5

f'(2) = 6(2) + 5 = 17

You need both coordinates of the point so f(2)= 3*2^2 + 5*2 - 2 = 20. So your point to use in the point-slope formula is (2,20) and your slope is 17

y-20 = 17(x- 2) or

y = 17x - 14

Normal means perpendicular

So the line that is perpendicular at the point (2,20) has a slope of -1/17, thus

y-20 = -1/17(x-2) or

y = (-1/17)x + 19 15/17

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