Exercise 11 Page 492

We are asked to graph the given quadratic function and find its zeroes. There are three steps that we need to do.

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

Since our function is already written in standard form, we can proceed by graphing it.

Graphing the Function

To draw the graph of the function written in standard form, we must start by identifying the values of $a,$ $b,$ and $c.$ We can see that $a=\text{-}1,$ $b=2,$ and $c=2.$ 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 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 of the parabola is the vertical line with equation $x=1.$

Calculating the Vertex

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 $x=1.$ Therefore, the $x\text{-}$coordinate of the vertex is also $1.$ To find the $y\text{-}$coordinate, we need to substitute $1$ for $x$ in the given equation. We found the $y\text{-}$coordinate, and now we know that the vertex is $(1,3).$

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

The $y\text{-}$intercept of the graph of a quadratic function written in standard form is given by the value of $c.$ Therefore, the point where our graph intercepts the $y\text{-}$axis is $(0,2).$ 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 $a=\text{-}1,$ which is negative, the parabola will open downwards. 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 function.

We can see that the parabola intersects the $x\text{-}$axis twice. The first point of intersection seems to be between $x=\text{-}1$ and $x=0,$ and the second one, between $x=2$ and $x=3.$ To approximate the zeroes, we will make tables using $x\text{-}$values using an increment of $0.1.$ Let's start with the table for $x$ between $\text{-}1$ and $0.$

$x$	$\text{-}0.9$	$\text{-}0.8$	$\text{-}0.7$	$\text{-}0.6$	$\text{-}0.5$	$\text{-}0.4$	$\text{-}0.3$	$\text{-}0.2$	$\text{-}0.1$
$f(x)$	$\text{-}0.61$	$\text{-}0.24$	$0.11$	$0.44$	$0.75$	$1.04$	$1.31$	$1.56$	$1.79$

As we can see, the change in signs happens between $x=\text{-}0.8$ and $x=\text{-}0.7,$ and the function value is closer to $0$ for $x=\text{-}0.7.$ Therefore, the first point of intersection is about $x=\text{-}0.7$. Let's make another table, this time for $x$ between $2$ and $3.$

$x$	$2.1$	$2.2$	$2.3$	$2.4$	$2.5$	$2.6$	$2.7$	$2.8$	$2.9$
$f(x)$	$1.79$	$1.56$	$1.31$	$1.04$	$0.75$	$0.44$	$0.11$	$\text{-}0.24$	$\text{-}0.61$

This time the change in signs happens between $x=2.7$ and $x=2.8,$ and the function value is closer to $0$ for $x=2.7.$ Therefore, the second point of intersection is about $x=2.7,$ so the zeroes of $f$ are about $\text{-}0.7$ and $2.7$