Question

If the line 2x-y=3 is tangent to the function f(x)=x²+2x-3, what is the point of tangency?

You must use a derivative to solve this problem

Answer 1

To find the point of tangency between the line 2x-y=3 and the function f(x)=x²+2x-3, set the expressions equal to each other and solve for x. The x-values correspond to the point of tangency, and substituting those values into the line equation gives the y-values. The point of tangency is (0, -3).

Step-by-step explanation:

To find the point of tangency between the line 2x-y=3 and the function f(x)=x²+2x-3, we need to find the x-coordinate of the point where they intersect. This can be done by setting the expression for the line equal to the expression for the function and solving for x.

First, let's rearrange the equation of the line to solve for y: y = 2x - 3.

Substitute this equation for y into the function equation: x² + 2x - 3 = 2x - 3.

Now we can solve for x: x² + 2x = 0. Factoring out x, we get x(x+2) = 0. This gives us two possible solutions: x = 0 and x = -2. Plugging these values back into the equation for the line, we get the two corresponding y-coordinates: y = 2(0) - 3 = -3 and y = 2(-2) - 3 = -7.

Therefore, the point of tangency is (0, -3).