find the slope of the tangent line to the parabola?

Question

x^2+2x point(-3,3)

also find an equation of the tangent line


please explain how to do it so i can learn, my professor does not explain even when asked, his response is to look it up in the book. please help thanks

it's basic to you because you know how to do it, i don't so it's not easy but thanks for explaining anyways.

to everyone else thank you very very much!

Answer 1

the derivative is 2x + 2. This is the formula for the slope of a tangent line at any point on the curve.
so plug in your x (-3)
2(-3) + 2
-6 + 2
-4
For the equation, use point-slope (-3,3) and m = -4
y - y1 = m(x - x1)
y - 3 = -4(x+4)
if you need it in slope-intercept form, get y by itself.

Answer 2

The slope of the tangent is going to be, by definition, the derivative. The derivative of x^2 + 2x is 2x+2, so the slope at x = -3, is -4 Now that you have the slope and a point on the line, you can use them to find the equation of the line.

Answer 3

slope = m = dy/dx = 2x+2
When x = -3 Slope = m = 2(-3)+2 = -4
So tangent is y=mx+b = -4x+b
When y= 3 , x = -3 so:
3 = -4(-3)+b
3 = 12+b
-9 = b
So tangent equation is y= -4x -9

Answer 4

I don't think you need a professor to learn something as basic as this...

Derivative = 2x + 2

Plug in the x value.

2(-3) + 2 = -6 + 2 = -4

Slope of Tangent Line = -4

y = -4x + b

Remember that it touched (-3,3)

3 = 12 + b

b = -9

y = -4x - 9
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Formula you'll use from today and maybe until the end of this week. You'll begin switching to using derivative rules soon.

lim h->0 [ f(x+h) - f(x) ] / h

To Geezah, yes I agree..
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And who gave thumbs down to my answer >.<.
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Now I get another?

Am I being hated or something?