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

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Answer 1 · 2018-04-15T10:39:55

This is in fact a quadratic in $y$

See the explanation

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

Square both sides

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

$x=1/3y^2-4/3$

Compare to $x=ay^2+by+c$

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

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

Determine the y-intercepts

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

$y=(0+-sqrt(0-4(1/3)(-4/3)))/(2(1/3))$

$y=+-sqrt(4^2/3^2)xx3/2$

$y=+-4/3xx3/2 = +-2$

Tony B

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