Solutions to Problem Set 30

Math 101-06

11-27-2017

[Graphing Quadratics]

1. Graph the quadratic function $y = x^2 - 2 x - 8$ . Find and label the roots, and find the x and y-coordinates of the vertex.

$$\eqalign{ x^2 - 2 x - 8 & = 0 \cr (x - 4)(x + 2) & = 0 \cr}$$

The roots are $x = 4$ and $x
   = -2$ .

The vertex is halfway between the roots, so the x-coordinate of the vertex is $x = \dfrac{4 +
   (-2)}{2} = 1$ . To find the y-coordinate, plug $x = 1$ into $y = x^2 - 2 x - 8$ :

$$y = 1^2 - 2 \cdot 1 - 8 = -9.$$

Since the coefficient of $x^2$ is positive, the graph opens upward.

$$\hbox{\epsfysize=1.5in \epsffile{graphing-quadratics1.eps}} \quad\halmos$$


2. Graph the quadratic function $y = -4 + 5x - x^2$ . Find and label the roots, and find the x and y-coordinates of the vertex.

$$\eqalign{ -4 + 5x - x^2 & = 0 \cr x^2 - 5x + 4 & = 0 \cr (x - 1)(x - 4) & = 0 \cr}$$

The roots are $x = 1$ and $x
   = 4$ .

The vertex is halfway between the roots, so the x-coordinate of the vertex is $x = \dfrac{1 + 4}{2}
   = 2.5$ . To find the y-coordinate, plug $x = 2.5$ into $y = -4 + 5x - x^2$ :

$$y = -4 + 5 \cdot \dfrac{5}{2} - \left(\dfrac{5}{2}\right)^2 = 2.25.$$

Since the coefficient of $x^2$ is negative, the graph opens downward.

$$\hbox{\epsfysize=1.5in \epsffile{graphing-quadratics2.eps}} \quad\halmos$$


3. Graph the quadratic function $y = x^2 - 4x + 4$ . Find and label the roots, and find the x and y-coordinates of the vertex.

$$\eqalign{ x^2 - 4x + 4 & = 0 \cr (x - 2)^2 & = 0 \cr}$$

The root is $x = 2$ .

The vertex is halfway between the roots, so the x-coordinate of the vertex is $x = \dfrac{2 + 2}{2}
   = 2$ . To find the y-coordinate, plug $x = 2$ into $y = x^2 -
   4x + 4$ :

$$y = 2^2 - 4 \cdot 2 + 4 = 0.$$

Since the coefficient of $x^2$ is positive, the graph opens upward.

$$\hbox{\epsfysize=1.5in \epsffile{graphing-quadratics3.eps}} \quad\halmos$$


4. Graph the quadratic function $y = 5x - x^2$ . Find and label the roots, and find the x and y-coordinates of the vertex.

$$\eqalign{ 5x - x^2 & = 0 \cr x(5 - x) & = 0 \cr}$$

The roots are $x = 0$ and $x
   = 5$ .

The vertex is halfway between the roots, so the x-coordinate of the vertex is $x = \dfrac{0 + 5}{2}
   = 2.5$ . To find the y-coordinate, plug $x = 2.5$ into $y = 5x - x^2$ :

$$y = 5 \cdot 2.5 - 2.5^2 = 6.25.$$

Since the coefficient of $x^2$ is negative, the graph opens downward.

$$\hbox{\epsfysize=1.5in \epsffile{graphing-quadratics4.eps}} \quad\halmos$$


5. Graph the quadratic function $y = x^2 - 4x + 5$ . Find and label the roots, and find the x and y-coordinates of the vertex.

Since $x^2 - 4x + 5$ doesn't factor, I'll find the roots using the Quadratic Formula:

$$x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - (4)(1)(5)}}{2 \cdot 1} = \dfrac{4 \pm \sqrt{-4}}{2} = \dfrac{4 \pm \sqrt{4}\sqrt{-1}}{2} = \dfrac{4 \pm 2i}{2} = 2 \pm i.$$

Since the roots are complex numbers, the graph doesn't cross the x-axis.

I'll use the formula $x =
   -\dfrac{b}{2a}$ to find the x-coordinate of the vertex; it is

$$x = -\dfrac{-4}{2 \cdot 1} = 2.$$

To find the y-coordinate, plug $x = 2$ into $y = x^2 - 4x + 5$ :

$$y = 2^2 - 4 \cdot 2 + 5 = 1.$$

Since the coefficient of $x^2$ is positive, the graph opens upward.

The only way the graph can open upward and not cross the x-axis is if it is "floating" above the x-axis:

$$\hbox{\epsfysize=1.5in \epsffile{graphing-quadratics5.eps}} \quad\halmos$$


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