Chapter 1.5, Problem 99E

To determine

three points which lie on the graph of the equation $y=-x^2+x-5$ .

Answer to Problem 99E

  $\left(0,-5\right)$ , $\left(1,-5\right)$ And $\left(2,-7\right)$

Explanation of Solution

Given:

The equation: $y=-x^2+x-5$

To find three points which lie on the graph of the equation $y=-x^2+x-5$ , substitute any three values for the variable x into the given equation and compute the corresponding y values.

First substitute $x=0$ in the given equation,

  $\begin{array}{c}y=-{\left(0\right)}^2+0-5\\ =0+0-5\\ =-5\end{array}$

So, one point on the graph is $\left(0,-5\right)$ .

Now substitute $x=1$ in the given equation,

  $\begin{array}{c}y=-{\left(1\right)}^2+1-5\\ =-1+1-5\\ =-5\end{array}$

So, another point on the graph is $\left(1,-5\right)$ .

Now substitute $x=2$ in the given equation,

  $\begin{array}{c}y=-{\left(2\right)}^2+2-5\\ =-4+2-5\\ =-7\end{array}$

So, another point on the graph is $\left(2,-7\right)$ .