If $f(x)=4 x^{2}-5 x+3$, find $f^{\prime}(1)$. Use this to find the equation of the tangent line to the parabola $y=4 x^{2}-5 x+3$ at the point $(1,2)$. The equation of this tangent line can be written in the form $y=m x+b$ where $m$ is: and where $b$ is:

Step 1 ：First, we need to find the derivative of the function \(f(x) = 4x^2 - 5x + 3\). The derivative of this function is \(f'(x) = 8x - 5\).

Step 2 ：Next, we substitute \(x = 1\) into the derivative to find the slope of the tangent line at the point \((1,2)\). This gives us \(f'(1) = 3\). So, the slope of the tangent line, \(m\), is 3.

Step 3 ：To find the y-intercept \(b\), we use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line. We know that the point \((1,2)\) is on the tangent line, so we substitute these values along with the slope \(m\) into the equation to solve for \(b\). This gives us \(b = 2 - 3(1 - 1) = 2/3\).

Step 4 ：Final Answer: The slope of the tangent line, \(m\), is \(\boxed{3}\) and the y-intercept, \(b\), is \(\boxed{\frac{2}{3}}\).