Exercise 9 Page 604

To draw the graph of the given quadratic function written in standard form, we must start by identifying the values of

a,

b,

and

c .

y = - \frac{1}{2} x^{2} + 4 x + 1 \Leftrightarrow y = - \frac{1}{2} x^{2} + 4 x + 1

We can see that

a = - \frac{1}{2},

b = 4,

and

c = 1 .

Now, we will follow four steps to graph the function.

Find the axis of symmetry.
Calculate the vertex.
Identify the $y -$ intercept and its reflection across the axis of symmetry.
Connect the points with a parabola.
Finding the Axis of Symmetry
The axis of symmetry is a vertical line with equation $x = - \frac{b}{2 a} .$ Since we already know the values of $a$ and $b,$ we can substitute them into the formula.
$x = - \frac{b}{2 a}$
SubstituteII
$a = - \frac{1}{2}$ , $b = 4$
$x = - \frac{4}{2 ( - \frac{1}{2} )}$

▼

Simplify right-hand side

MultPosNeg
$a (- b) = - a \cdot b$
$x = - \frac{4}{- 1}$
MoveNegDenomToFrac
Put minus sign in front of fraction
$x = - (- \frac{4}{1})$
NegNeg
$- (- a) = a$
$x = \frac{4}{1}$
DivByOne
$\frac{a}{1} = a$

$x = 4$
The axis of symmetry of the parabola is the vertical line with equation $x = 4 .$
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 $x -$ and $y -$ coordinates in terms of $a$ and $b .$
$Vertex : (- \frac{b}{2 a}, f (- \frac{b}{2 a}))$
Note that the formula for the $x -$ coordinate is the same as the formula for the axis of symmetry, which is $x = 4 .$ Thus, the $x -$ coordinate of the vertex is also $4 .$ To find the $y -$ coordinate, we need to substitute $4$ for $x$ in the given equation.
$y = - \frac{1}{2} x^{2} + 4 x + 1$
Substitute
$x = 4$
$y = - \frac{1}{2} (4)^{2} + 4 (4) + 1$

▼

Simplify right-hand side

CalcPow
Calculate power
$y = - \frac{1}{2} \cdot 16 + 4 (4) + 1$
Multiply
Multiply
$y = - 8 + 16 + 1$
AddTerms
Add terms

$y = 9$
We found the $y -$ coordinate, and now we know that the vertex is $(4, 9) .$
Identifying the $y -$ intercept and its Reflection

The $y -$ intercept of the graph of a quadratic function written in standard form is given by the value of $c .$ Thus, the point where our graph intercepts the $y -$ axis is $(0, 1) .$ 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 $a = - \frac{1}{2},$ which is negative, the parabola will open downwards. Let's connect the three points with a smooth curve.

Exercise 9 Page 604

Finding the Axis of Symmetry

Calculating the Vertex

Identifying the y-intercept and its Reflection

Connecting the Points

Identifying the $y -$ intercept and its Reflection