Finding C?

[donayre21]

Hi my problem is a graph of a parabola y=x^2
They show a line going through the parabola.Point (1,1) is in the parabola,point (0,2) is in the middle not touching any thing beside the line going through(this is the mid point of the line).The last point is point c who is touching the last left side of the parabola.A line is going through all 3 points including the unknown C.I have tried the midpoint formula to find C put I have not been successful as the answer is (-2,4).I'm just really confused on why that is the answer, when using the mid point formula should word as the mid point and the first point were given.I'm sorry if what I wrote does not make any sense. Thank you for the help and I look forward to the explanation of why the answer for c is (-2,4).

 

[fcabanski]

Hi my problem is a graph of a parabola y=x^2
They show a line going through the parabola.Point (1,1) is in the parabola,point (0,2) is in the middle not touching any thing beside the line going through(this is the mid point of the line).The last point is point c who is touching the last left side of the parabola.A line is going through all 3 points including the unknown C.I have tried the midpoint formula to find C put I have not been successful as the answer is (-2,4).I'm just really confused on why that is the answer, when using the mid point formula should word as the mid point and the first point were given.I'm sorry if what I wrote does not make any sense. Thank you for the help and I look forward to the explanation of why the answer for c is (-2,4).

The problem doesn't specify c is the midpoint. You're assuming it by looking at the diagram.

You can find the slope of the line given the two points. Plug that slope, and one of the points, into the standard equation for a line. y-y(1) = m(x-x(1)) where (x(1), y(1)) is one of the given points.

Set that equation = to the equation for the parabola, because you want to find points where the equations intersect. Solve for x. It will result in two answers, but one is the other (known) point, where the graphs intersect. Plug that back into the parabola equation or the line equation to find the corresponding y value.

 

[donayre21]

Thank you for the help it actually helps.I just have one more question I got this equation x^2+x-2 which gave me two answers x=1 and x=-2.The -2 is one of the point right and I can find y by plugging it in to the parabola equation y=x^2? You said in your reply that I could also plug it in into the linear equation,how would you actually do that because I get x^2-y1=-x-2 once I have plugged in the new point to y-y1=m(x-x1).One more thing will the information that you gave me work if any of the other point were missing as long as I have 2 points given?Thanks again

 

[donayre21]

for par b of the problem I'm given the excact sam picture but now with point (0,b) and (1,1).I tried doing it the same way but end up with b-1 as my slope and when I use the linear equation just as I did in the first one it takes me no where.

 

[HallsofIvy]

So you still have the parabola \(\displaystyle y= x^2\), a line through (1, 1) and (0, b), and want to determine the other point at which the line crosses the parabola?

The slope of the line is \(\displaystyle \frac{1- b}{1- 0}= 1- b\), NOT b- 1. So the equation of the line is y- 1= (1- b)(x- 1)
or y= (1- b)x- (1- b)+ 1= (1-b)x+ b. Replacing y in \(\displaystyle y= x^2\) by that we have \(\displaystyle (1- b)x+ b= x^2\), which reduces to \(\displaystyle x^2+ (b- 1)x- b= 0\). Knowing that x= 1 is a solution (because (1, 1) is on both the line and parabola) we know that can be factored as (x- 1)(x- ?) and so determine the other x where the line crosses the parabola.

 

[donayre21]

So you still have the parabola \(\displaystyle y= x^2\), a line through (1, 1) and (0, b), and want to determine the other point at which the line crosses the parabola?

The slope of the line is \(\displaystyle \frac{1- b}{1- 0}= 1- b\), NOT b- 1. So the equation of the line is y- 1= (1- b)(x- 1)
or y= (1- b)x- (1- b)+ 1= (1-b)x+ b. Replacing y in \(\displaystyle y= x^2\) by that we have \(\displaystyle (1- b)x+ b= x^2\), which reduces to \(\displaystyle x^2+ (b- 1)x- b= 0\). Knowing that x= 1 is a solution (because (1, 1) is on both the line and parabola) we know that can be factored as (x- 1)(x- ?) and so determine the other x where the line crosses the parabola.

Alright so I got up to here,x^2-x-bx+b,because I multiplied (1-b)(x-1) and used the parabola equation.I would have just taken the equation that I got apart but I have 4 terms instead of 3.This is were I'm really confused on because on the problem before I unfoiled and got two numbers,one of them was not the answer as it was already in the line and to get the y intercept I plugged my x to my equation for the parabola Y=X^2.How would I unfoil this 4 term equation?

 

[HallsofIvy]

The quadratic equation you get is \(\displaystyle x^2+ (b-1)x- b= 0\). As I said, we already know that x= 1 is a solution because the line and parabola intersect at x= 1. So (x- 1) is a factor and we must have (x- 1)(x-?). That will give \(\displaystyle x^2- x- ?x+ ?= x^2+ (b- 1)x- b\) so comparing the last terms on each side, we must have ?= -b. \(\displaystyle (x- 1)(x+ b)= x^2- x+ bx- b= x^2+ (b- 1)x- b\). x= 1 and x= -b are the two places where the line and parabola cross.