Chapter 2.1, Problem 93E

To determine

To find:The point of intersections of graphs of given equations; algebraically, and verify the results using intersect feature of a graphing utility.

Answer to Problem 93E

The point of intersection of the curve $y=x^3+2x-1$ , and the line $y=-2x+15$ is $\left(x,y\right)\equiv \left(2,11\right)$ .

Explanation of Solution

Given data:

The given equations are $y=x^3+2x-1$ and $y=-2x+15$ .

Method used:

For algebraic solution (point of intersection) of the equations, we use back substitution method.

The givenequations are

  $y=x^3+2x-1$...............(1)

  $y=-2x+15$...............(2)

From equation (2) substituting $y=-2x+15$ in equation (1)

  $-2x+15=x^3+2x-1$

  $\Rightarrow x^3+4x-16=0$

  $\Rightarrow x^3-2x^2+2x^2-4x+8x-16=0$

  $\Rightarrow x^2\left(x-2\right)+2x\left(x-2\right)+8\left(x-2\right)$

  $\Rightarrow \left(x-2\right)\left(x^2+2x+8\right)=0$

  $\Rightarrow \left(x-2\right)=0,\left(x^2+2x+8\right)=0$

  $\Rightarrow \left(x-2\right)=0\Rightarrow x=2$ , and $x^2+2x+8=0$

  $\Rightarrow x=\frac{-2\pm \sqrt{{\left(2\right)}^2-4\times 1\times 8}}{2}=\frac{-2\pm \sqrt{4-32}}{2}=-1\pm \sqrt{-7}$ , which is an imaginary number.

  $\Rightarrow x=2\Rightarrow y=11$ .

Thus, the point of intersection of the equations (1) and (2) is $\left(x,y\right)\equiv \left(2,11\right)$ .

The graphs of equations are $y=x^3+2x-1$ and $y=-2x+15$ are shown in Figure-1 here.

  

From the graph, it is clear that the point of intersection of the curve $y=x^3+2x-1$ , and theline $y=-2x+15$ is $\left(x,y\right)\equiv \left(2,11\right)$ that verify the results obtained algebraically.

Conclusion:

The point of intersection of the curve $y=x^3+2x-1$ , and the line $y=-2x+15$ is $\left(x,y\right)\equiv \left(2,11\right)$ .