Exercise 1 Page 490

We are asked to solve the given quadratic equation by graphing. There are three steps to solving a quadratic equation by graphing.

1. Write the equation in standard form, $a{x}^2+bx+c=0.$
2. Graph the related function $y=a{x}^2+bx+c.$
3. Find the $x\text{-}$intercepts, if any.

The solutions of $a{x}^2+bx+c=0$ are the $x\text{-}$intercepts of the graph of $y=a{x}^2+bx+c.$ Our equation is already written in standard form. Let's identify the function related to the equation.

Graphing the Related Function

To draw the graph of the related function written in standard form, we must start by identifying the values of $a,$ $b,$ and $c.$ We can see that $a=1,$ $b=\text{-}1,$ and $c=\text{-}2.$ Now, we will follow four steps to graph the function.

1. Find the axis of symmetry.
2. Calculate the vertex.
3. Identify the $y\text{-}$intercept and its reflection across the axis of symmetry.
4. Connect the points with a parabola.

Finding the Axis of Symmetry

The axis of symmetry is a vertical line with the equation $x=\text{-}\frac{b}{2a}.$ Since we already know the values of $a$ and $b,$ we can substitute them into the formula. The axis of symmetry for the parabola is a vertical line with the equation $x=0.5.$

Calculating the Vertex

To calculate the vertex, we need to think of $y$ as a function of $x,$ $y=f(x).$ We can write the expression for the vertex by stating the $x\text{-}$ and $y\text{-}$coordinates in terms of $a$ and $b.$ Note that the formula for the $x\text{-}$coordinate is the same as the formula for the axis of symmetry, which is $x=0.5.$ Therefore, the $x\text{-}$coordinate of the vertex is also $0.5.$ To find the $y\text{-}$coordinate, we need to substitute $0.5$ for $x$ in the given equation. We found the $y\text{-}$coordinate and now we know that the vertex is $(0.5,\text{-}2.25).$

Identifying the $y\text{-}$intercept and its Reflection

The $y\text{-}$intercept of the graph for a quadratic function written in standard form is given by the value of $c.$ The point where our graph intercepts the $y\text{-}$axis is $(0,\text{-}2).$ Let's plot this point and its reflection across the axis of symmetry.

Connecting the Points

We can now draw the graph of the function. Since $a=1,$ which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.

Finding the $x\text{-}$intercepts

Let's identify the $x\text{-}$intercepts of the graph of the related function.

We can see that the parabola intersects the $x\text{-}$axis twice. The points of intersection are $(\text{-}1,0)$ and $(2,0).$ Therefore, the equation $x^2-x-2=0$ has two solutions, $x=\text{-}1$ and $x=2.$