## pottersheep Group Title 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 one year ago one year ago

1. pottersheep Group Title

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

2. satellite73 Group Title

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 Group Title

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

4. satellite73 Group Title

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

5. pottersheep Group Title

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

6. satellite73 Group Title

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

7. satellite73 Group Title

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

8. pottersheep Group Title

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

9. satellite73 Group Title

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

10. pottersheep Group Title

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 Group Title

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

12. satellite73 Group Title

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

13. satellite73 Group Title

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

14. satellite73 Group Title

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 Group Title

I get it now!!!

16. pottersheep Group Title

Thank you so much :)

17. satellite73 Group Title

ok good yw

18. pottersheep Group Title

You are such a good teacher =D