Graph f(x)=x^2-4x+32 ; at (3,0) , t^3

$f (x) = x^{2} - 4 x + 32$ ; at $(3, 0)$ , $t^{3}$

Step 1

Graph $f (x) = x^{2} - 4 x + 32$ .

Find the properties of the given parabola.

Rewrite the equation in vertex form.

Complete the square for $x^{2} - 4 x + 32$ .

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = - 4$

$c = 32$

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 4}{2 \cdot 1}$

Cancel the common factor of $- 4$ and $2$ .

Factor $2$ out of $- 4$ .

$d = \frac{2 \cdot - 2}{2 \cdot 1}$

Cancel the common factors.

Factor $2$ out of $2 \cdot 1$ .

$d = \frac{2 \cdot - 2}{2 (1)}$

Cancel the common factor.

$d = \frac{2 \cdot - 2}{2 \cdot 1}$

Rewrite the expression.

$d = \frac{- 2}{1}$

Divide $- 2$ by $1$ .

$d = - 2$

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 32 - \frac{{(- 4)}^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Cancel the common factor of ${(- 4)}^{2}$ and $4$ .

Rewrite $- 4$ as $- 1 (4)$ .

$e = 32 - \frac{{(- 1 (4))}^{2}}{4 \cdot 1}$

Apply the product rule to $- 1 (4)$ .

$e = 32 - \frac{{(- 1)}^{2} \cdot 4^{2}}{4 \cdot 1}$

Raise $- 1$ to the power of $2$ .

$e = 32 - \frac{1 \cdot 4^{2}}{4 \cdot 1}$

Multiply $4^{2}$ by $1$ .

$e = 32 - \frac{4^{2}}{4 \cdot 1}$

Factor $4$ out of $4^{2}$ .

$e = 32 - \frac{4 \cdot 4}{4 \cdot 1}$

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

$e = 32 - \frac{4 \cdot 4}{4 (1)}$

Cancel the common factor.

$e = 32 - \frac{4 \cdot 4}{4 \cdot 1}$

Rewrite the expression.

$e = 32 - \frac{4}{1}$

Divide $4$ by $1$ .

$e = 32 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 32 - 4$

Subtract $4$ from $32$ .

$e = 28$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - 2)}^{2} + 28$ .

${(x - 2)}^{2} + 28$

Set $y$ equal to the new right side.

$y = {(x - 2)}^{2} + 28$

Use the vertex form, $y = a {(x - h)}^{2} + k$ , to determine the values of $a$ , $h$ , and $k$ .

$a = 1$

$h = 2$

$k = 28$

Since the value of $a$ is positive, the parabola opens up.

Opens Up

Find the vertex $(h, k)$ .

$(2, 28)$

Find $p$ , the distance from the vertex to the focus.

Find the distance from the vertex to a focus of the parabola by using the following formula.

$\frac{1}{4 a}$

Substitute the value of $a$ into the formula.

$\frac{1}{4 \cdot 1}$

Cancel the common factor of $1$ .

Cancel the common factor.

$\frac{1}{4 \cdot 1}$

Rewrite the expression.

$\frac{1}{4}$

Find the focus.

The focus of a parabola can be found by adding $p$ to the y-coordinate $k$ if the parabola opens up or down.

$(h, k + p)$

Substitute the known values of $h$ , $p$ , and $k$ into the formula and simplify.

$(2, \frac{113}{4})$

Find the axis of symmetry by finding the line that passes through the vertex and the focus.

$x = 2$

Find the directrix.

The directrix of a parabola is the horizontal line found by subtracting $p$ from the y-coordinate $k$ of the vertex if the parabola opens up or down.

$y = k - p$

Substitute the known values of $p$ and $k$ into the formula and simplify.

$y = \frac{111}{4}$

Use the properties of the parabola to analyze and graph the parabola.

Direction: Opens Up

Vertex: $(2, 28)$

Focus: $(2, \frac{113}{4})$

Axis of Symmetry: $x = 2$

Directrix: $y = \frac{111}{4}$

Direction: Opens Up

Vertex: $(2, 28)$

Focus: $(2, \frac{113}{4})$

Axis of Symmetry: $x = 2$

Directrix: $y = \frac{111}{4}$

Select a few $x$ values, and plug them into the equation to find the corresponding $y$ values. The $x$ values should be selected around the vertex.

Replace the variable $x$ with $1$ in the expression.

$f (1) = {(1)}^{2} - 4 \cdot 1 + 32$

Simplify the result.

Simplify each term.

One to any power is one.

$f (1) = 1 - 4 \cdot 1 + 32$

Multiply $- 4$ by $1$ .

$f (1) = 1 - 4 + 32$

Simplify by adding and subtracting.

Subtract $4$ from $1$ .

$f (1) = - 3 + 32$

Add $- 3$ and $32$ .

$f (1) = 29$

The final answer is $29$ .

$29$

The $y$ value at $x = 1$ is $29$ .

$y = 29$

Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{2} - 4 \cdot 0 + 32$

Simplify the result.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 - 4 \cdot 0 + 32$

Multiply $- 4$ by $0$ .

$f (0) = 0 + 0 + 32$

Simplify by adding numbers.

Add $0$ and $0$ .

$f (0) = 0 + 32$

Add $0$ and $32$ .

$f (0) = 32$

The final answer is $32$ .

$32$

The $y$ value at $x = 0$ is $32$ .

$y = 32$

Replace the variable $x$ with $3$ in the expression.

$f (3) = {(3)}^{2} - 4 \cdot 3 + 32$

Simplify the result.

Simplify each term.

Raise $3$ to the power of $2$ .

$f (3) = 9 - 4 \cdot 3 + 32$

Multiply $- 4$ by $3$ .

$f (3) = 9 - 12 + 32$

Simplify by adding and subtracting.

Subtract $12$ from $9$ .

$f (3) = - 3 + 32$

Add $- 3$ and $32$ .

$f (3) = 29$

The final answer is $29$ .

$29$

The $y$ value at $x = 3$ is $29$ .

$y = 29$

Replace the variable $x$ with $4$ in the expression.

$f (4) = {(4)}^{2} - 4 \cdot 4 + 32$

Simplify the result.

Simplify each term.

Raise $4$ to the power of $2$ .

$f (4) = 16 - 4 \cdot 4 + 32$

Multiply $- 4$ by $4$ .

$f (4) = 16 - 16 + 32$

Simplify by adding and subtracting.

Subtract $16$ from $16$ .

$f (4) = 0 + 32$

Add $0$ and $32$ .

$f (4) = 32$

The final answer is $32$ .

$32$

The $y$ value at $x = 4$ is $32$ .

$y = 32$

Graph the parabola using its properties and the selected points.

$\begin{array}{c} x & y \\ 0 & 32 \\ 1 & 29 \\ 2 & 28 \\ 3 & 29 \\ 4 & 32 \end{array}$

Graph the parabola using its properties and the selected points.

Direction: Opens Up

Vertex: $(2, 28)$

Focus: $(2, \frac{113}{4})$

Axis of Symmetry: $x = 2$

Directrix: $y = \frac{111}{4}$

$\begin{array}{c} x & y \\ 0 & 32 \\ 1 & 29 \\ 2 & 28 \\ 3 & 29 \\ 4 & 32 \end{array}$

Direction: Opens Up

Vertex: $(2, 28)$

Focus: $(2, \frac{113}{4})$

Axis of Symmetry: $x = 2$

Directrix: $y = \frac{111}{4}$

$\begin{array}{c} x & y \\ 0 & 32 \\ 1 & 29 \\ 2 & 28 \\ 3 & 29 \\ 4 & 32 \end{array}$

Step 2

Graph $t^{3}$ .

Find the point at $x = - 2$ .

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = {(- 2)}^{3}$

Simplify the result.

Raise $- 2$ to the power of $3$ .

$f (- 2) = - 8$

The final answer is $- 8$ .

$- 8$

Convert $- 8$ to decimal.

$y = - 8$

Find the point at $x = - 1$ .

Replace the variable $x$ with $- 1$ in the expression.

$f (- 1) = {(- 1)}^{3}$

Simplify the result.

Raise $- 1$ to the power of $3$ .

$f (- 1) = - 1$

The final answer is $- 1$ .

$- 1$

Convert $- 1$ to decimal.

$y = - 1$

Find the point at $x = 0$ .

Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{3}$

Simplify the result.

Raising $0$ to any positive power yields $0$ .

$f (0) = 0$

The final answer is $0$ .

$0$

Convert $0$ to decimal.

$y = 0$

Find the point at $x = 1$ .

Replace the variable $x$ with $1$ in the expression.

$f (1) = {(1)}^{3}$

Simplify the result.

One to any power is one.

$f (1) = 1$

The final answer is $1$ .

$1$

Convert $1$ to decimal.

$y = 1$

Find the point at $x = 2$ .

Replace the variable $x$ with $2$ in the expression.

$f (2) = {(2)}^{3}$

Simplify the result.

Raise $2$ to the power of $3$ .

$f (2) = 8$

The final answer is $8$ .

$8$

Convert $8$ to decimal.

$y = 8$

The cubic function can be graphed using the function behavior and the points.

$\begin{array}{c} x & y \\ - 2 & - 8 \\ - 1 & - 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 8 \end{array}$

The cubic function can be graphed using the function behavior and the selected points.

Falls to the left and rises to the right

$\begin{array}{c} x & y \\ - 2 & - 8 \\ - 1 & - 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 8 \end{array}$

Falls to the left and rises to the right

$\begin{array}{c} x & y \\ - 2 & - 8 \\ - 1 & - 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 8 \end{array}$

Step 3

Plot each graph on the same coordinate system.

$f (x) = x^{2} - 4 x + 32$

$(3, 0)$

$t^{3}$

Step 4

Graph f(x)=x^2-4x+32 ; at (3,0) , t^3

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