What is the Gradient of the Tangent to a Curve at a Given Point?

[DizzyDoo]

[SOLVED] Basic Differentiation

Hi there, I've heard that this forum is the best for getting some quick help.

Find the gradient of the tangent to the curve at the given point:
y = 2x² + 3x + 5 when x = -1

Homework Equations

Right, I'm very new a this, but dy/dx is the way to go right?

3. The Attempt at a Solution :
y = 2x² + 3x + 5 when x = -1
dy/dx it, and I get:
dy/dx=4x + 3
Substitute -1 in and I get
y=-1

So I've now got two points; (-1,-1) but that's not the gradient, but I've got the feeling I'm half way there. Any help would be great thanks!

 

[HallsofIvy]

Hi there, I've heard that this forum is the best for getting some quick help.

Find the gradient of the tangent to the curve at the given point:
y = 2x² + 3x + 5 when x = -1

Homework Equations

Right, I'm very new a this, but dy/dx is the way to go right?

3. The Attempt at a Solution :
y = 2x² + 3x + 5 when x = -1
dy/dx it, and I get:
dy/dx=4x + 3
Substitute -1 in and I get
y=-1

So I've now got two points; (-1,-1) but that's not the gradient, but I've got the feeling I'm half way there. Any help would be great thanks!

Well, first of all (-1, -1) is ONE point, not two! But more important, if dy/dx= 4x+ 3, then putting x=-1 gives you dy/dx= -1, not y. At first I thought that it was a typo but then you say "that's not the gradient". It certainly IS the gradient! You are not "half way there", you are completely there!

(It is true that, if x=-1, y= 2(-1)²+ 3(-1)+ 5= 2- 3+ 5= 4. y= 4 so the graph of the function passes through the point (-1, 4). The tangent line to the curve is y= -1(x+1)+ 4 or y= -x+ 3.)

 

[DizzyDoo]

Fantastic, I can finish off the rest of the questions now. Thanks!