Question

Find the slope of the tangent line to the graph of the function at the given point. \(g(x)=\frac{3}{2} x+1, \quad(-2,-2)\)

Short answer

The slope of the tangent line to the function \(g(x)=\frac{3}{2} x+1\) at the point (-2,-2) is \(\frac{3}{2}\).

Step-by-step solution

Identify the slope of the function

For the function \(g(x)=\frac{3}{2} x+1\), the slope (which is the physical interpretation of the derivative for a linear function) is \(\frac{3}{2}\) which is the coefficient of \(x\). This is the same regardless of the point on the line, as the function is linear.

Confirm the point lies on the function

Before claiming the slope of the tangent line at the point (-2,-2), check to ensure this point actually lies on the function. Substitute the coordinates of the point into the original function. With \(x=-2\), the function's value becomes \(\frac{3}{2}*(-2)+1=-3+1=-2\). Since the result is -2 which equals to the y-coordinate of the given point, the point (-2,-2) does indeed lie on the function.

State the slope of the tangent line at the given point

The slope of the tangent line at any point on the function \(g(x)\) equals the derivative which is the coefficient of \(x\), and that derivative or slope is \(\frac{3}{2}\)