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Question: How do you use the graphing utility to graph \(f(x) = - 2{x^2} + 10x\) and identify the \(x\) interc...

How do you use the graphing utility to graph f(x)=−2x2+10xf(x) = - 2{x^2} + 10x and identify the xx intercepts and the vertex?

Explanation

Solution

This problem deals with the conic sections. A conic section is a curve obtained as the intersection of the surface of a cone with a plane. There are three such types of conic sections which are, the parabola, the hyperbola and the ellipse. This problem is regarding one of those conic sections, which is a parabola. The general form of an equation of a parabola is given by x2=−4ay{x^2} = - 4ay.

Complete step-by-step solution:
Here consider the given parabola of equation f(x)=−2x2+10xf(x) = - 2{x^2} + 10x
Here let the function f(x)f(x) be yy, as given below:
⇒y=f(x)\Rightarrow y = f(x)
So the equation of the parabola is given by:
⇒y=−2x2+10x\Rightarrow y = - 2{x^2} + 10x
Taking the term 2x2x common on the right hand side of the above equation as given below:
⇒y=−2x(x−5)\Rightarrow y = - 2x\left( {x - 5} \right)
To find the xx intercepts, put y=0y = 0, as given below:
⇒−2x(x−5)=0\Rightarrow - 2x\left( {x - 5} \right) = 0
∴x=0;x=5\therefore x = 0;x = 5
So the xx intercepts are 0 and 5.
The graph of the given parabola is shown below:

Now consider the given parabola equation y=−2x2+10xy = - 2{x^2} + 10x, writing this in its standard form as shown below:
⇒(x−2.5)2=−12(y−12.5)\Rightarrow {\left( {x - 2.5} \right)^2} = - 12\left( {y - 12.5} \right)
If the above equation is simplified, the given equation of parabola y=−2x2+10xy = - 2{x^2} + 10x, will be obtained.
Here in the above equation, it shows that the vertex VV is :
⇒V=(2.5,12.5)\Rightarrow V = \left( {2.5,12.5} \right)
This parabola has its axis parallel to y-axis.

The xx intercepts are 0 and 5, whereas the vertex is (2.5,12.5)\left( {2.5,12.5} \right).

Note: Please note that if the given parabola is x2=−4ay{x^2} = - 4ay, then the vertex of this parabola is the origin (0,0)\left( {0,0} \right), and there is no intercept for this parabola as there are no terms of x or y. If the equation of the parabola includes any terms of linear x or y, then the vertex of the parabola is not the origin, the vertex has to be found by simplifying it into its particular standard form.