Question

find the equation of the tangent line to the graph of f at the indicated value of x. f(x)=15x^{2}+6x; x = 0 y=

Explanation:

Step1: Find the derivative of \(f(x)\)

The derivative of \(f(x)=15x^{2}+6x\) using the power - rule \((x^n)^\prime=nx^{n - 1}\) is \(f^\prime(x)=30x + 6\).

Step2: Evaluate the derivative at \(x = 0\)

Substitute \(x = 0\) into \(f^\prime(x)\): \(f^\prime(0)=30\times0+6=6\). This is the slope \(m\) of the tangent line.

Step3: Find the \(y\) - value of the function at \(x = 0\)

Substitute \(x = 0\) into \(f(x)\): \(f(0)=15\times0^{2}+6\times0 = 0\). So the point \((x_0,y_0)=(0,0)\).

Step4: Use the point - slope form \(y - y_0=m(x - x_0)\)

Substitute \(m = 6\), \(x_0 = 0\) and \(y_0 = 0\) into the point - slope form \(y-0=6(x - 0)\), we get \(y=6x\).

Answer:

\(y = 6x\)