1) Find the point on the graph of y=x^2 where the cube has slope 5/2

2) Find the point on the graph of y=x^2 where the tangent line is parallel to the line -4x+3y=3

The first question has a type point should be on the curve of the graph not the cube

since the slope anywere is just y'=2x, you want to find where

2x = 5/2
x = 5/4
So, the desired point is (5/4,25/16)

For the other one, figure the slope of the tangent line, and then follow the same steps.

1) To find the point on the graph of y = x^2 where the derivative (slope) is equal to 5/2, we need to calculate the derivative of the function y = x^2, set it equal to 5/2, and solve for x.

First, let's find the derivative of y = x^2. The power rule states that if y = x^n, then the derivative dy/dx = n * x^(n-1). Applying this rule to y = x^2, we have:

dy/dx = 2 * x^(2-1)
dy/dx = 2 * x

Now we have the derivative of y = x^2, which represents the slope of the tangent line at any point on the graph.

Setting dy/dx = 5/2, we can solve for x:

2 * x = 5/2
x = 5/4

Now that we have the x-coordinate, we can substitute it into the original equation y = x^2 to find the y-coordinate:

y = (5/4)^2
y = 25/16

Therefore, the point on the graph of y = x^2 where the slope is 5/2 is (5/4, 25/16).

2) To find the point on the graph of y = x^2 where the tangent line is parallel to the line -4x + 3y = 3, we need to find the derivative of y = x^2, set it equal to the derivative of the given line, and solve for x.

First, let's find the derivative of y = x^2 using the power rule:

dy/dx = 2 * x

Now, the given line -4x + 3y = 3 needs to be rearranged in slope-intercept form (y = mx + b) to determine the slope. Let's solve for y:

-4x + 3y = 3
3y = 4x + 3
y = (4/3)x + 1

The derivative of this line represents its slope, which is 4/3.

Now we set the derivative of y = x^2 equal to the slope of the given line:

2 * x = 4/3

Solving for x:

x = (4/3) / 2
x = 2/3

Now that we have the x-coordinate, we can substitute it into the original equation y = x^2 to find the y-coordinate:

y = (2/3)^2
y = 4/9

Therefore, the point on the graph of y = x^2 where the tangent line is parallel to the line -4x + 3y = 3 is (2/3, 4/9).