# How do you graph the quadratic function and identify the vertex and axis of symmetry and x intercepts for y=1/3(x+4)(x+1)?

Jul 24, 2018

The vertex is $\left(- \frac{5}{2} , - \frac{3}{4}\right)$ or $\left(- 2.5 , - 0.75\right)$.
The y-intercept is $\left(0 , \frac{4}{3}\right)$ or $\left(0 , \approx 1.333\right)$.
The x-intercepts are $\left(- 1 , 0\right) , \left(- 4 , 0\right)$.
Additional point: $\left(- 5 , \frac{4}{3}\right)$ or $\left(- 5 , \approx 1.333\right)$.

#### Explanation:

Given:

$y = \frac{1}{3} \left(x + 4\right) \left(x + 1\right)$

Expand $\left(x + 4\right) \left(x + 1\right)$.

$y = \frac{1}{3} \left({x}^{2} + 5 x + 4\right)$

Distribute $\frac{1}{3}$.

$y = \frac{1}{3} {x}^{2} + \frac{5}{3} x + \frac{4}{3}$ is a quadratic equation in standard form:

$y = a {x}^{2} + b x + c$,

where:

$a = \frac{1}{3}$, $b = \frac{5}{3}$, and $c = \frac{4}{3}$

The vertex is the maximum or minimum point of the parabola. The formula for the axis of symmetry gives us the x-coordinate of the vertex:

$x = \frac{- b}{2 a}$

$x = \frac{- \frac{5}{3}}{2 \cdot \frac{1}{3}}$

$x = \frac{- \frac{5}{3}}{\frac{2}{3}}$

$x = - \frac{5}{3} \times \frac{3}{2}$

$x = - \frac{15}{6}$

$x = - \frac{5}{2}$ or $2.5$

To find the y-coordinate of the vertex, substitute $- \frac{5}{2}$ for $x$ and solve for $y$.

$y = \frac{1}{3} {\left(- \frac{5}{2}\right)}^{2} + \frac{5}{3} \left(- \frac{5}{2}\right) + \frac{4}{3}$

$y = \frac{1}{3} \left(\frac{25}{4}\right) - \frac{25}{6} + \frac{4}{3}$

$y = \frac{25}{12} - \frac{25}{6} + \frac{4}{3}$

The least common denominator is $12$. Multiply $\frac{25}{6} \times \frac{2}{2}$ and $\frac{4}{3} \times \frac{4}{4}$ to get equivalent fractions. Since $\frac{n}{n} = 1$, the numbers will change but the value of each fraction will not change.

$y = \frac{25}{12} - \frac{25}{6} \times \frac{\textcolor{red}{2}}{\textcolor{red}{2}} + \frac{4}{3} \times \frac{\textcolor{b l u e}{4}}{\textcolor{b l u e}{4}}$

$y = \frac{25}{12} - \frac{50}{12} + \frac{16}{12}$

$y = - \frac{9}{12}$

$y = - \frac{3}{4}$ or $- 0.75$

The vertex is $\left(- \frac{5}{2} , - \frac{3}{4}\right)$ or $\left(- 2.5 , - 0.75\right)$. Plot this point.

The y-intercept is the value of $y$ when $x = 0$. Substitute $0$ for $x$ and solve for $y$.

$y = \frac{1}{3} {\left(0\right)}^{2} + \frac{5}{3} \left(0\right) + \frac{4}{3}$

$y = \frac{4}{3}$ or $\approx 1.333$

The y-intercept is $\left(0 , \frac{4}{3}\right)$ or $\left(0 , \approx 1.333\right)$. Plot this point.

The x-intercepts are the values for $x$ when $y = 0$. Substitute $0$ for $y$ and solve for $x$.

$0 = \frac{1}{3} {x}^{2} + \frac{5}{3} x + \frac{4}{3}$

Switch sides.

$\frac{1}{3} {x}^{2} + \frac{5}{3} x + \frac{4}{3} = 0$

Multiply both sides by $3$.

${x}^{2} + 5 x + 4 = 0$

Factor ${x}^{2} + 5 x + 4$.

$\left(x + 1\right) \left(x + 4\right) = 0$

Set each binomial to zero and solve.

$x + 1 = 0$

$x = - 1$

$x + 4 = 0$

$x = - 4$

The x-intercepts are $\left(- 1 , 0\right) , \left(- 4 , 0\right)$. Plot these points.

Additional point: $x = - 5$

$x = - 5$ is the mirror of the x-coordinate of the y-intercept.

Substitute $- 5$ for $x$ and solve for $y$.

$y = \frac{1}{3} {\left(- 5\right)}^{2} + \frac{5}{3} \left(- 5\right) + \frac{4}{3}$

$y = \frac{25}{3} - \frac{25}{3} + \frac{4}{3}$

Additional point: $\left(- 5 , \frac{4}{3}\right)$ or $\left(- 5 , \approx 1.333\right)$. Plot this point.

Sketch a graph through the points. Do not connect the dots.

graph{y=(x^2)/3+(5x)/3+4/3 [-10, 10, -5, 5]}