Chapter 4.7, Problem 1E

To determine

To graph: the quadratic equation as mentioned in the problem.

Explanation of Solution

Given information:

The quadratic equation is $y=-x^2+2x-1$ .

Graph:

The graph mentioned below is in real number line of x axis and y axis.

Where vertex $\left(1,0\right)$ .

  

Interpretation:

Consider the quadratic equation $y=-x^2+2x-1$ .

The mentioned quadratic equation above will be in vertex form $y=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex.

We can write the mentioned quadratic equation in the vertex form as $y=-x^2+2x-1\Rightarrow y=-\left(x^2-2x+1\right)\Rightarrow y=-{\left(x-1\right)}^2.$

Therefore the vertex form of the equation is $y=-{\left(x-1\right)}^2+0$ .

Which is comparable to $y=a{\left(x-h\right)}^2+k$ where $a=-1$ which is less than $0$ .

Since $a<0$ the direction of graph will be negative side of y axis and the vertex is $\left(1,0\right)$ .

The discriminant of standard form of quadratic equation $y=-x^2+2x-1$ is $D=b^2-4*a*c\Rightarrow D={\left(2\right)}^2-4*\left(-1\right)*\left(-1\right)=4-4=0.$

Here $D=0$ .

Therefore the graph will lie on x axis.

At $x=0$ the $y=-x^2+2x-1$ will yield $y=-1$ .

Therefore graph will cut y axis at $\left(0,-1\right)$ .

Therefore the graph will be as mentioned below.

  

To determine

To draw: the table for the equation $y=-x^2+2x-1$ .

Answer to Problem 1E

The table of the equation is

  $\begin{array}{cc}x& y\\ -3& -16\\ -2& -9\\ -1& -4\\ 0& -1\\ 1& 0\\ 2& -1\\ 3& -4\end{array}$ .

Explanation of Solution

Given information:

The equation is $y=-x^2+2x-1$ and the value of $x$ is $-3,-2,-1,0,1,2,3.$

Formula:

We will substitute the value of $x$ in the equation $y=-x^2+2x-1$ .

Calculation:

Consider the equation $y=-x^2+2x-1$ .

Since the value of $x$ is $-3,-2,-1,0,1,2,3.$

Recall that we will substitute the value of $x$ in the equation $y=-x^2+2x-1$ .

Therefore

At $x=-3\Rightarrow y=-\left(-3*-3\right)+2\left(-3\right)-1=-16.$

At $x=-2\Rightarrow y=-\left(-2*-2\right)+2\left(-2\right)-1=-9.$ .

At $x=-1\Rightarrow y=-\left(-1*-1\right)+2\left(-1\right)-1=-4.$ .

At $x=0\Rightarrow y=-\left(0*0\right)+2\left(0\right)-1=-1.$ .

At $x=1\Rightarrow y=-\left(1*1\right)+2\left(1\right)-1=0.$

At $x=2\Rightarrow y=-\left(2*2\right)+2\left(2\right)-1=-1.$

At $x=3\Rightarrow y=-\left(3*3\right)+2\left(3\right)-1=-4.$

Therefore the table will be as mentioned below.

  $\begin{array}{cc}x& y\\ -3& -16\\ -2& -9\\ -1& -4\\ 0& -1\\ 1& 0\\ 2& -1\\ 3& -4\end{array}$