An old HSC question

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AlonzoN

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The equation of the tangent to the curve y= x^3 + ax^2+ bx + 4 at the point where x = 2 is y = x - 4.

Find the values of "a" and "b".

So what I did was differentiated
 
Then what did you do?

Seeing you have 2 unknowns, you need to form 2 simultaneous equations.
 
The equation of the tangent to the curve y= x^3 + ax^2+ bx + 4 at the point where x = 2 is y = x - 4.

Find the values of "a" and "b".

So what I did was differentiated

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

What you need to to do know is

y - y1 = m(x - x1)

And what I know is that when you plug the "x" value into the first equation is that it will give you the y-value, and if you put the x-value into the differentiated equation, it will give you the gradient.

So this is what I did:

y = y1 and y' = m, so

y - (12 + 4a +2b) = (12 + 4a + b)*(x-2)

and that got me,

y - 12 - 4a - 2b = 4ax - 8a + bx - 2b + 12x - 24

simplify that,

y + 12 + 4a = 4ax + bx + 12x

And now I'm stuck.
 
Don't forget that when x=2, that is where the tangent and the cubic meet. So not only does y = 12a + 4a +2b but it also equals what?

Also, the gradient at that point is equal to the gradient of the tangent which is ???
 
Harry_the_cat said:
Don't forget that when x=2, that is where the tangent and the cubic meet. So not only does y = 12a + 4a +2b but it also equals what?

Also, the gradient at that point is equal to the gradient of the tangent which is ???
Click to expand...
So are you saying that y is also = 12 + 4a + b?

And the gradient at that point is equal to 2. So would I need to plug "2", as the gradient?
 
No, When x=2, the point on the cubic graph also lies on the tangent line. Yes?
 
NO, you're not quite getting it. May I suggest you draw a rough cubic function (roughly like 2 parabolas joined) with a max and a min.. Pick a point and draw a tangent line. Trust me, it will help. It doesn't have to be accurate - just a squiggle and a tangent line.
 
YES! y=-2 because that point lies one the tangent line y=x-4. Good!
Before you said when x=2, y= 12 + 4a +2b which is also correct, because the point also lies on the cubic.

SO..... what equals what?
 
Harry_the_cat said:
YES! y=-2 because that point lies one the tangent line y=x-4. Good!
Before you said when x=2, y= 12 + 4a +2b which is also correct, because the point also lies on the cubic.

SO..... what equals what?
Click to expand...
OH!!!! -2 = 12 + 4a + 2b, right? So then I can just make a = 0, work out b, vice versa, and then try it out on a graphing calculator, right?
 
Yes -2 = 12 + 4a +2b. This is one equation with two variables You need two equations.

Why would you put a=0? You need to find a and b. You can't just make it 0. So forget that idea!

We need to find another equation with a and b and solve simultaneously.

Let's look at gradients.
What is the gradient at the point where x=2?
You had said it before.
 
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