find the eqt. of tangent to the curve y=-x^2+2x-10 @ the point where the curve cuts the y-axis.[ans:y=2x-10]
how do i do this because they didn't give me the point...?
Yes, they did
It said, "where the curve cuts the y-axis", that is when x = 0
when x=0, y = -0 + 0 - 10 = -10
so your point is (0,-10)
drwls answered this for you already, did you not read it?
http://www.jiskha.com/display.cgi?id=1311337279
To find the equation of the tangent to the curve at the point where it cuts the y-axis, we first need to determine the coordinates of that point.
When a curve cuts the y-axis, it means that the x-coordinate of that point is 0. So, to find the point where the curve cuts the y-axis, we substitute x = 0 into the equation of the curve:
y = -x^2 + 2x - 10
Substituting x = 0:
y = -(0)^2 + 2(0) - 10
y = 0 + 0 - 10
y = -10
Therefore, the point where the curve cuts the y-axis is (0, -10). Now that we have the point, we can proceed to find the equation of the tangent.
To find the equation of the tangent, we need two pieces of information: the slope of the tangent line and a point on the line.
The slope of the tangent line can be found by taking the derivative of the curve's equation with respect to x.
Differentiating y = -x^2 + 2x - 10 with respect to x:
dy/dx = -2x + 2
Now, substitute x = 0 (the x-coordinate of the point on the tangent line) into the derivative:
m = -2(0) + 2
m = 2
So, the slope of the tangent line is 2.
Now, we have both the slope of the tangent line (m = 2) and a point on the line (0, -10). We can use the point-slope form of the equation of a line to derive the equation of the tangent line:
y - y1 = m(x - x1)
Substituting the values:
y - (-10) = 2(x - 0)
y + 10 = 2x
Simplifying the equation, we get:
y = 2x - 10
Therefore, the equation of the tangent to the curve y = -x^2 + 2x - 10 at the point where it cuts the y-axis is y = 2x - 10.