## pottersheep 2 years ago Please help. Calculus! Find the point of tangency and the equation of the tangent line passing through the point (-1.-23) to the graph of the function f(x) = 3 + 2x + 3x^2

1. pottersheep

The answer says:q (-4,43) y = -22x - 45 AND y = 14x - 9 at q(2,19)

2. satellite73

the derivative is $$f'(x)=2-6x$$ the slope of the tangent line through $$(x,2+2x+3x^2)$$ and $$(-1,-23)$$ is $\frac{2+2x+2x^2+23}{x+1}$

3. satellite73

set $\frac{2+2x+2x^2+23}{x+1}=2+6x$and solve for $$x$$

4. satellite73

oops, i meant the derivative is $$f'(x)=2+6x$$

5. pottersheep

Thank you for your time. Sorry, why is it (2+2x+2x2+23)/x+1? Where did the x+1 come from?

6. satellite73

slope is $$\frac{y_2-y_1}{x_2-x_1}$$

7. satellite73

here you have the point $$(-1,-23)$$ so the denominator is $$x+1$$

8. pottersheep

What exactly does the point P(-1,-23) mean?

9. satellite73

it says that $$(-1,-23)$$ is a point on the tangent line, so it is given to you

10. pottersheep

What does it give me? Is it because a tangent is a straight line so I can use that point and the slope formula?

11. pottersheep

Ah, I thought it was where the tangent met f(x) = 3 + 2x + 3x^2....but that didnt make sense to me

12. satellite73

a generic point on the curve looks lik e $(x,3+2x+3x^2)$ you also have the point $$(-1,-23)$$

13. satellite73

the slope of the line through those two points is $\frac{3+2x+2x^2+23}{x+1}$

14. satellite73

and you know that the line with this slope is tangent to the curve at some point the slope my therefore be equal to the derivative at that point, which is $$2+6x^2$$

15. pottersheep

I get it now!!!

16. pottersheep

Thank you so much :)

17. satellite73

ok good yw

18. pottersheep

You are such a good teacher =D