Chapter 9.1, Problem 5AGP

To determine

To graph the given function.

Explanation of Solution

Given information :

Given function is $f\left(x\right)=-2x^2+2x-1$ .

Graph :

Interpretation :

The graph of a quadratic equation ( $y=a{x}^2+bx+c$ ) is a parabola. The parabola is symmetric in nature and the point of symmetry is known as the vertex. This vertex lies on the line of symmetry and is symmetric about this line.

This vertex point shall be:

Highest point (if $a<0$ ) and be called the maximum. Here, the graph opens downwards.

A parabola always points to infinity, either negative or positive.

To graph a quadratic function, compute the axis of symmetry, vertex and y-intercept, post which, plot the same on a graph.

The axis of symmetry bisects the parabola into two equal parts. Hence each point on the parabola would have an equal point on the other side of the axis of symmetry. Plot these points on the graph with a smooth curve.

Formula to compute equation of the axis of symmetry $x=-\frac{b}{2\times a}$ . Here, $a=-2$ and $b=2$ . ‘a’ and ‘b’ are coefficients $x^2$ and $x$ respectively.

Axis of symmetry for the given function is

  $x=-\frac{b}{2\times a}$

Formula for axis of symmetry.

  $=-\frac{2}{2\times \left(-2\right)}$

Putting the values of ‘a’ and ‘b’ .

  $=\frac{1}{2}$

  $x=\frac{1}{2}$ is the equation of axis of symmetry.

Vertex can be found out by putting the value of x computed in the axis of symmetry in the original function. This will give a value of y . These two coordinates of x and y would be the point where the vertex is.

  $=-2{\left(\frac{1}{2}\right)}^2+2\left(\frac{1}{2}\right)-1$

Putting the value of $x=\frac{1}{2}$ in the original function.

  $=-2\times \frac{1}{4}+1-1$

Simplifying the expression.

  $=-\frac{1}{2}$

Thus the vertex is $(\frac{1}{2},-\frac{1}{2})$ .

y-intercept is computed by substituting the value of x in the equation by 0.

  $=-2{\left(0\right)}^2+2\left(0\right)-1$

Simplifying

  $-1$

Hence the point of y-intercept is $\left(0,-1\right)$ .

Now, plot these points along with their reflecting symmetric points, starting from the vertex.