Parabola graph problem

Question

Find the maximum vertical distance between parabola and the line Actual answer is 6.25 But why we can't just count it from graph but it give us 6

Answer 1

Here's what I get:

If you denote the parabola $f_1(x)$ and the linear function $f_2(x)$ you can specify the distance between the two, at least in the area marked in the figure, as:

$y(x) = f_1(x) - f_2(x) = -2x^2+4x+3-x+2=-2x^2+3x+5$

Derivate to find at what value of $x$ the maximum distance occurs:

$\frac{dy}{dx}=-4x+3=0 \Rightarrow x = \frac{3}{4}$

Insert into expression for distance again:

$y(\frac{3}{4}) = -2\frac{9}{16}+4\frac{3}{4}+3-\frac{3}{4}+2 = 6.125$

Answer 2

Because the max distance is at the point when the tangent to the parabola is parallel to the line. The slope of the tangent line to parabola at $x_0$ is: $$f'(x_0)=-4x_0+4=1 \Rightarrow x_0=\frac{3}{4}.$$ Hence the distance between the tangent and the line at $x_0$ is: $$d=\left(-2\cdot \left(\frac34\right)^2+4\cdot\frac34+3\right)-\left(\frac34-2\right)=6.125.$$