Graph x=3t and y = square root of t+6 , 0<=t<=8

$x = 3 t$ and $y = \sqrt{t} + 6$ , $0 \leq t \leq 8$

Graph $x = 3 t$ .

Solve for $y$ .

Rewrite the equation as $3 y = x$ .

$3 y = x$

Divide each term in $3 y = x$ by $3$ and simplify.

Divide each term in $3 y = x$ by $3$ .

$\frac{3 y}{3} = \frac{x}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{x}{3}$

Divide $y$ by $1$ .

$y = \frac{x}{3}$

Rewrite in slope-intercept form.

The slope-intercept form is $y = m x + b$ , where $m$ is the slope and $b$ is the y-intercept.

$y = m x + b$

Reorder terms.

$y = \frac{1}{3} x$

Use the slope-intercept form to find the slope and y-intercept.

Find the values of $m$ and $b$ using the form $y = m x + b$ .

$m = \frac{1}{3}$

$b = 0$

The slope of the line is the value of $m$ , and the y-intercept is the value of $b$ .

Slope: $\frac{1}{3}$

y-intercept: $(0, 0)$

Slope: $\frac{1}{3}$

y-intercept: $(0, 0)$

Any line can be graphed using two points. Select two $x$ values, and plug them into the equation to find the corresponding $y$ values.

Reorder terms.

$y = \frac{1}{3} x$

Create a table of the $x$ and $y$ values.

$\begin{array}{c} x & y \\ 0 & 0 \\ 3 & 1 \end{array}$

Graph the line using the slope and the y-intercept, or the points.

Slope: $\frac{1}{3}$

y-intercept: $(0, 0)$

$\begin{array}{c} x & y \\ 0 & 0 \\ 3 & 1 \end{array}$

Slope: $\frac{1}{3}$

y-intercept: $(0, 0)$

$\begin{array}{c} x & y \\ 0 & 0 \\ 3 & 1 \end{array}$

Graph $y = \sqrt{t} + 6$ .

Find the domain for $y = \sqrt{t} + 6$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

Interval Notation:

$[0, \infty)$

Set-Builder Notation:

${x | x \geq 0}$

To find the radical expression end point, substitute the $x$ value $0$ , which is the least value in the domain, into $f (x) = \sqrt{x} + 6$ .

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

$f (0) = \sqrt{0} + 6$

Simplify the result.

Remove parentheses.

$f (0) = \sqrt{0} + 6$

Simplify each term.

Rewrite $0$ as $0^{2}$ .

$f (0) = \sqrt{0^{2}} + 6$

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = 0 + 6$

Add $0$ and $6$ .

$f (0) = 6$

The final answer is $6$ .

$6$

The radical expression end point is $(0, 6)$ .

$(0, 6)$

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

Substitute the $x$ value $1$ into $f (x) = \sqrt{x} + 6$ . In this case, the point is $(1, 7)$ .

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

$f (1) = \sqrt{1} + 6$

Simplify the result.

Remove parentheses.

$f (1) = \sqrt{1} + 6$

Any root of $1$ is $1$ .

$f (1) = 1 + 6$

Add $1$ and $6$ .

$f (1) = 7$

The final answer is $7$ .

$y = 7$

Substitute the $x$ value $2$ into $f (x) = \sqrt{x} + 6$ . In this case, the point is $(2, \sqrt{2} + 6)$ .

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

$f (2) = \sqrt{2} + 6$

Simplify the result.

Remove parentheses.

$f (2) = \sqrt{2} + 6$

The final answer is $\sqrt{2} + 6$ .

$y = \sqrt{2} + 6$

The square root can be graphed using the points around the vertex $(0, 6), (1, 7), (2, 7.41)$

$\begin{array}{c} x & y \\ 0 & 6 \\ 1 & 7 \\ 2 & 7.41 \end{array}$

Graph $0 \leq t \leq 8$ .

$0 \leq t \leq 8$

Plot each graph on the same coordinate system.

$x = 3 t$

$y = \sqrt{t} + 6$

$0 \leq t \leq 8$

Graph x=3t and y = square root of t+6 , 0<=t<=8

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