How do you find the vertex, focus and sketch $y^{2} = 3 x$ $y^2=3x$ ?

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Answer 1 · 2017-05-17T08:41:43

Vertex: $(0, 0)$ $(0, 0)$
Directrix: $x = - 3 / 4$ $x=-3//4$
Focus: $(\frac{3}{4}, 0)$ $(3/4, 0)$

When the $y$ $y$ term is squared, the parabola opens to the left (or right). When a parabola can be expressed as

$y^{2} = 4 p x$ $y^2=4px$

The vertex is at $(0, 0)$ $(0, 0)$

The directrix is at $x = - p$ $x=-p$

The focus is at $(p, 0)$ $(p, 0)$

The given equation, $y^{2} = 3 x$ $y^2=3x$ , is almost there, but the number in front of $x$ $x$ doesn't have a $4$ $4$ in there. What do we do?

We want the $3 = 4 p$ $3=4p$ . Notice if I divide both sides of this equation aby $4$ $4$ , you get

$p = \frac{3}{4}$ $p=3/4$

This allows us to rewrite the original as

$y^{2} = 4 (\frac{3}{4}) x$ $y^2=4(3/4)x$

Therefore,
Vertex: $(0, 0)$ $(0, 0)$
Directrix: $x = - 3 / 4$ $x=-3//4$
Focus: $(\frac{3}{4}, 0)$ $(3/4, 0)$

graph{(y^2-3x)(x+3/4)=0 [-1.2, 2, -3, 3]}

How do you find the vertex, focus and sketch y^2=3xy2=3xy^2=3x?