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enter image description here 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

Marva Jami
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2 Answers2

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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: $$d=\left(-2\cdot \left(\frac34\right)^2+4\cdot\frac34+3\right)-\left(\frac34-2\right)=6.125.$$

farruhota
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  • If 5 is maximum value of parabola how can it be possible to have 6.25? – Marva Jami Feb 10 '18 at 11:52
  • The question is to find max vertical distance between parabola and the line. $5$ is irrelevant. Think of moving the green line up as fas as it touches the parabola. You can draw two lines to compare distances: $y=x+4$ and $y=x+\frac{43}{8}$. Actual answer must be $6.125$. – farruhota Feb 10 '18 at 13:55
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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:

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

tahaum
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