Need help with Pre-Cal Question?

Question

Find a point (a,b) on the curve y=x^2 such that the line through the point (a,b) and the point (1,1) has slop 1. Express your answer as a point.

Please help me and detail how you got your answer so I can take notes! Best answer will be given!

Answer 1

m=1
We are trying to find a point (x,y).
Since this point will be on the graph of y=x^2, we can say that the point is at (x, x^2) because y=x^2. Now, we can find (change in y)/(change in x)=1 between the 2 points.

(1-x^2)/(1-x)=1
Cross multiply
1-x=1-x^2
Simplify
x^2-x=0
x(x-1)=0
x=0, 1

Plug each of our values for x into y=x^2 to find our y values.

y=(1)^2 and y=(0)^2

When x=1, y=1
When x=0, y=0.

We have 2 answers, but since one of the answers is the point (1,1), we have to disregard it because we are trying to find a line connecting 2 points, (x,y) and (1,1). So, our answer is the point (0,0).

I hope that this helps.

Answer 2

To solve this question, recall the slope formula:

m = (y2 - y1) / (x2 - x1)

We want to find a point on the curve y = x^2 (which we call (a,b) ) such that the point (a,b) and (1,1) has slope 1.

Note that if (a,b) is on the curve, it follows that since y = x^2,
b = a^2, so the point can also be written as (a, a^2).

Using the slope formula, with (1,1) as (x1, y1) and (a,a^2) as (x2,y2), we have:

m = (a^2 - 1) / (a - 1)

But, we're given that our slope m = 1, so

1 = (a^2 - 1) / (a - 1)

We can factor the numerator on the right hand side as a difference of squares.

1 = [(a - 1)(a + 1)] / (a - 1)

Which leads to a cancellation

1 = a + 1

Therefore, a = 0. It follows that b = 0^0 = 0.

Therefore, the point on the curve y = x^2 that has a slope of 1 with (1,1) is (0,0).

Answer 3

No algebra is needed to obtain the required point. ***

Consider even a crude sketch of the parabola y = x^2.

The point (1, 1) happens to lie on it. A line with slope 1 (or 45 degrees) through (1, 1) clearly passes through the origin at (0, 0), and THAT is ALSO a point on the curve. It's necessarily the unique solution to the question posed, since a line can only meet a quadratic curve in two points at most. We're done! :

The point is (0, 0). QED.

Live long and prosper.

*** P.S. All too often, algebraic methods lead to far too long and tedious solutions. Graphical or geometrical approaches are often shorter, and certainly more elegant.

Answer 4

Both (a,b) and (1,1) are on y=x^2
The line through the point (1,1) with slope 1 is
y= mx+b = x+b1 = 1 +b so b=0
The line is y=x
This line intersects the parabola y= x^2 at (0, 0) and (1,1) and has a slope of 1 so either of these two points satisfy the problem.

Answer 5

A point of the curv has coordinates(x,x^2) Give two points of a line the slope is: slope =(x2-1)/(x-1) = x+1
So x+1= 1 x=0 y=0 solution(0,0)
I put x instead of"a"

Answer 6

Pre-Cal ? That's plain co-ordinate geometry !!!!!!!!!!