Question

(a) find the slope of the tangent to the parabola y=3x^2+2x at the point whose x-coordinate is a.
(b) at what point on the parabola is the tangent line parallel to the line y=10x-2?
thanks!

Answer 1

can u use derivatives well?

Answer 2

i don't know derivatives yet. i just start doing calculus on limits

Answer 3

i got part a), that answer is 6a+2, but I'm not sure how to do part b)

Answer 4

so you found y'=6x+2
at y' at x=a you have 6a+2
so the slope of the tangent line at x=a is 6a+2
-
So question b tells you to find at what point on the parabola the tangent line is paralel to y=10x-2
We know parallel lines have the same slope (and different y-intercept)
so what is the slope of y=10x-2?

Answer 5

And you will set 6a+2=slope of(10x-2)
and solve for a

Answer 6

a=5/3x-2/3?

Answer 7

the slope of 10x-2 is?

Answer 8

\[6a+2=\text{ slope of }(10x+2) \\ 6a+2=10\]

Answer 9

10x-2 whatever

Answer 10

but still slope of 10x-2 is 10

Answer 11

@esam2 Do you understand that the slope of the tangent line has to be the same as the other line if we want them to be parallel?

Answer 12

yes

Answer 13

a=4/3

Answer 14

slope of tangent line at x=a is 6a+2
slope of the other line is 10

we want to find when those slopes are the same
6a+2 is the same as 10 when a=?
6a=10-2
6a=8
a=8/6=4/3
yep
now it says to find the point
you found the x value such that the slopes are the same
now find the y on the parabola for that given x

Answer 15

f(x)=3x^2+2x
so
f(4/3)=?

Answer 16

is it 8? so it will be (4/3,8)?

Answer 17

\[f(\frac{4}{3})=3(\frac{4}{3})^2+2(\frac{4}{3}) \\ 3 \cdot \frac{4}{3} \cdot \frac{4}{3}+\frac{8}{3} \\ \frac{16}{3}+\frac{8}{3}=\frac{24}{3}=8\]
yeah

Answer 18

thanks!

Answer 19

np