How do you graph $y = \sqrt{3 x + 4}$ $y=sqrt(3x+4)$ ?

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How do you graph $y = \sqrt{3 x + 4}$ $y=sqrt(3x+4)$ ?

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Answer 1 · 2018-04-15T10:39:55

This is in fact a quadratic in $y$ $y$

See the explanation

Explanation:

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

Square both sides

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

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

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

$There is no b y term so the axis of symmetry is the x-axis$ $color(green)("There is no "by" term so the axis of symmetry is the x-axis")$

$The x-intercept (vertex) is at x = - \frac{4}{3}$ $color(green)("The x-intercept (vertex) is at "x=-4/3)$

Determine the y-intercepts

$y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $y=(-b+-sqrt(b^2-4ac))/(2a)$ where $a = \frac{1}{3}, b = 0 and c = - \frac{4}{3}$ $a=1/3, b=0 and c=-4/3$

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

$y = \pm \sqrt{\frac{4^{2}}{3^{2}}} \times \frac{3}{2}$ $y=+-sqrt(4^2/3^2)xx3/2$

$y = \pm \frac{4}{3} \times \frac{3}{2} = \pm 2$ $y=+-4/3xx3/2 = +-2$

Tony B

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How do you graph $y = \sqrt{3 x + 4}$ $y=sqrt(3x+4)$ ?

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Explanation:

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How do you graph y=√3x+4y=sqrt(3x+4)?

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Explanation:

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How do you graph $y = \sqrt{3 x + 4}$ $y=sqrt(3x+4)$ ?