An old HSC question

y' = 3x^2 + 2ax + b

when x = 2,

y = 2^3 + 2^2a + 2b + 4
= 12 + 4a + 2b

and

y' = 3*2^2 + 2*2a + b
= 12 + 4a + b

This, yes?
 
Oh, right, y' is the gradient at x = 2, so would I need to plug -2 as y and then solve 12 + 4a + b?
 
When you said 2 equations, could I do b = -12 - 4a, and then plug that into y?
 
I put -2 into "y" and got

-2 = 12 + 4a + 2b

-14 = 4a + 2b

-7 = 2a + b

so then I tried to make b the subject,

b = -2a - 7

plugged that into y',

y' = 12 + 4a - 2a - 7
= 5 + 2a

but I'm guessing that isn't it, is it?
 
The equation of the tangent line is y= 1x-4. This line has a slope of 1. This means that f'(2) = 1
Recall, that the derivative at x=2 gives the slope of the tangent line (which is 1)

f'(x) = 3x^2 +2ax + b. Then, f'(2) = 12 + 4a + b = 1

So one equation is 4a + b = -11

When x = 2, y= x - 4 = -2. So one point on the line is (2, -2). But this point is also on y = x^3 + ax^2+ bx + 4. That is, -2 = 8 + 4a + 2b + 4. This gives us a 2nd equation of 2a + b = -7

Solve this system of equations for a and b.
 
So what do I do?

Simultaneous equations?

make b the subject for one of them, then plug b into the other?

like:

b = -7 -2a

so,

4a -7 - 2a = -11

2a = -4

therefore, a = -2, then do I plug "a" into one of them, or both just to be safe?

4*-2 + b = -11, therefore b = -3,

2*-2 + b = -7, therefore b = -3.

So a = -2 and b = -3. Did I do it right?
 
I just tried it out on a graphing calculator, and it works, THANK YOU SO MUCH!!!!!
 
AlonzoN said:
So what do I do?

Simultaneous equations?

make b the subject for one of them, then plug b into the other?

like:

b = -7 -2a

so,

4a -7 - 2a = -11

2a = -4

therefore, a = -2, then do I plug "a" into one of them, or both just to be safe?

4*-2 + b = -11, therefore b = -3,

2*-2 + b = -7, therefore b = -3.

So a = -2 and b = -3. Did I do it right?
Click to expand...
Looks good to me. You should check that both equations are valid for the results you got for a and b.
 
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