Show that the tangent to the parabola Y=Ax^2 (for A does not equal 0) at the point where x = c will intersect the x axis at the point (c/2 , 0 . Where does it intersect the y axis.
To find the tangent find the derivative of $\displaystyle y=ax^2$ at $\displaystyle x=c$ which is $\displaystyle \lim_{h\rightarrow 0} \frac{a(x+h)^2-ax^2}{h}=2ac$Originally Posted by frozenflames
Thus, the slope of the tangent line is $\displaystyle m=2ac$. Now use the formula for finding equations of line by knowing thier slopes and the point that they pass through. That formula is $\displaystyle y-y_0=m(x-x_0)$this tangent line has slope $\displaystyle m=2ac$ and passes through the point of tangentcy $\displaystyle x=c$ thus $\displaystyle y=ac^2$ Thus, the equation of the tangent line is $\displaystyle y-ac^2=2ac(x-c)$ thus, $\displaystyle y=2acx-2ac^2+ac^2=2acx-ac^2$ now set $\displaystyle y=0$ because this is the x-intercept thus $\displaystyle 2acx-ac^2=0$ solve for $\displaystyle x$ thus$\displaystyle x=c/2$
Q.E.D.