find the eqt. of tangent to the curve y=-x^2+2x-10.[ans:y=2x-10]
but my ans is y=-9. if im wrong could you show the calculation.
At what point along the curve is the tangent contact made?
Based upon the answer, it is where dy/dx = 2
dy/dx = -2x + 2 = 2, so x = 0 at the point of contact. That requires y = -10 at that point.
So y = 2x - 10 is the answer IF you are talking about the line tangent to the curve at x = 0.
You omitted an important part of the problem statement.
in the question it say: at the point where the curve cuts the y-axis
To find the equation of the tangent to the curve given by y = -x^2 + 2x - 10, we can use the concept of differentiation.
First, differentiate the curve equation with respect to x to find the derivative (dy/dx):
dy/dx = -2x + 2
The derivative represents the rate of change of y with respect to x. Now, let's find the value of x for which the slope of the tangent is equal to the derivative. Since the slope of the tangent is equal to dy/dx, we can set -2x + 2 equal to the desired slope of 2:
-2x + 2 = 2
Solving this equation will give us the x-coordinate where the tangent line intersects the curve. Subtracting 2 from both sides:
-2x = 0
This simplifies to x = 1.
Now, substitute this value of x back into the original equation to find the corresponding y-coordinate:
y = -(1)^2 + 2(1) - 10
= -1 + 2 - 10
= -9
So, at the point (1, -9), the tangent line intersects the curve.
Using the slope-intercept form of a line (y = mx + c), we can now find the equation of the tangent line by substituting the coordinates of the point (1, -9):
y = mx + c
-9 = 2(1) + c
-9 = 2 + c
c = -9 - 2
c = -11
Therefore, the equation of the tangent line is y = 2x - 11, not y = 2x - 10.
Apologies for the confusion earlier.