# How do you graph y=sqrt(3x+4)?

Apr 15, 2018

This is in fact a quadratic in $y$

See the explanation

#### Explanation:

Given: $y = \sqrt{3 x + 4}$

Square both sides

${y}^{2} = 3 x + 4$ Now we are off!

$x = \frac{1}{3} {y}^{2} - \frac{4}{3}$

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

$\textcolor{g r e e n}{\text{There is no "by" term so the axis of symmetry is the x-axis}}$

$\textcolor{g r e e n}{\text{The x-intercept (vertex) is at } x = - \frac{4}{3}}$

Determine the y-intercepts

$y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ where $a = \frac{1}{3} , b = 0 \mathmr{and} c = - \frac{4}{3}$

$y = \frac{0 \pm \sqrt{0 - 4 \left(\frac{1}{3}\right) \left(- \frac{4}{3}\right)}}{2 \left(\frac{1}{3}\right)}$

$y = \pm \sqrt{{4}^{2} / {3}^{2}} \times \frac{3}{2}$

$y = \pm \frac{4}{3} \times \frac{3}{2} = \pm 2$