Zerrotolerance
New member
- Joined
- Oct 3, 2010
- Messages
- 20
I know a similar question was asked a few posts up, but this one is a little different and I can't seem to figure it out. The problem is:
Find a parabola of the form given below that has slope m[sub:qp8n34yo]1[/sub:qp8n34yo] at x[sub:qp8n34yo]1[/sub:qp8n34yo], slope m[sub:qp8n34yo]2[/sub:qp8n34yo] at x[sub:qp8n34yo]2[/sub:qp8n34yo], and passes through the point P.
y = ax[sup:qp8n34yo]2[/sup:qp8n34yo] + bx + c
m[sub:qp8n34yo]1[/sub:qp8n34yo] = 12 , x[sub:qp8n34yo]1[/sub:qp8n34yo]= 8
m[sub:qp8n34yo]2[/sub:qp8n34yo] = 7, x[sub:qp8n34yo]2[/sub:qp8n34yo] = 4
P= (4,3)
My first thought when I saw this was to find the derivitive of y. I did this using the rules of finding derivitives for constants, powers, sums, and differences. (Don't know if I am saying that right, but hopefully you know what I mean.) I came up with:
y' = 2ax + b
The derivitive of x[sup:qp8n34yo]2[/sup:qp8n34yo]is 2x, so that gives 2ax.
The derivitive of x to the first power is 1, so that leaves b(1).
c is a constant and the derivitive of a constant is 0, so it was dropped.
From here I know that the derivitive of y is the slope for a point if I plug in x. What's getting me is I don't know what to do to find the constants a,b,c. I seem to be drawing a blank. One thing that I did notice is that the slope is 7 for the point x[sub:qp8n34yo]2[/sub:qp8n34yo]= 4. It wants the equation of the line that goes through point (4,3). So the slope of that point should be 7 because it seems to be the same point they tell you about above. So, I know how to find the equation of the tangent line to the point, but not the equation of the parabola.
I have tried just about everything I can think of. I plugged 4 in for x and 3 in for y in the original equation and tried coming up with an a,b, and c that works, but when I plug them into y' to see if they work, they don't. Can someone give me a clue as to what I need to do next to figure this out? Thanks!
Find a parabola of the form given below that has slope m[sub:qp8n34yo]1[/sub:qp8n34yo] at x[sub:qp8n34yo]1[/sub:qp8n34yo], slope m[sub:qp8n34yo]2[/sub:qp8n34yo] at x[sub:qp8n34yo]2[/sub:qp8n34yo], and passes through the point P.
y = ax[sup:qp8n34yo]2[/sup:qp8n34yo] + bx + c
m[sub:qp8n34yo]1[/sub:qp8n34yo] = 12 , x[sub:qp8n34yo]1[/sub:qp8n34yo]= 8
m[sub:qp8n34yo]2[/sub:qp8n34yo] = 7, x[sub:qp8n34yo]2[/sub:qp8n34yo] = 4
P= (4,3)
My first thought when I saw this was to find the derivitive of y. I did this using the rules of finding derivitives for constants, powers, sums, and differences. (Don't know if I am saying that right, but hopefully you know what I mean.) I came up with:
y' = 2ax + b
The derivitive of x[sup:qp8n34yo]2[/sup:qp8n34yo]is 2x, so that gives 2ax.
The derivitive of x to the first power is 1, so that leaves b(1).
c is a constant and the derivitive of a constant is 0, so it was dropped.
From here I know that the derivitive of y is the slope for a point if I plug in x. What's getting me is I don't know what to do to find the constants a,b,c. I seem to be drawing a blank. One thing that I did notice is that the slope is 7 for the point x[sub:qp8n34yo]2[/sub:qp8n34yo]= 4. It wants the equation of the line that goes through point (4,3). So the slope of that point should be 7 because it seems to be the same point they tell you about above. So, I know how to find the equation of the tangent line to the point, but not the equation of the parabola.
I have tried just about everything I can think of. I plugged 4 in for x and 3 in for y in the original equation and tried coming up with an a,b, and c that works, but when I plug them into y' to see if they work, they don't. Can someone give me a clue as to what I need to do next to figure this out? Thanks!