If $f(x)=4x^2-5x+3$, find $f^{\prime }(1)$.
Use this to find the equation of the tangent line to the parabola $y=4x^2-5x+3$ at the point $(1,2)$. The equation of this tangent line can be written in the form $y=mx+b$ 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 $f^{\prime }(x)=8x-5$.

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 $f^{\prime }(1)=3$. 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 $\boxed{{\displaystyle 3}}$ and the y-intercept, $b$, is $\boxed{{\displaystyle \frac{2}{3}}}$.