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Question: write an equation for each parabola with vertex at the origin

Problem: through (3,2); symmetric with respect to the x axis

Respuesta :

Step-by-step explanation:

Since the axis of symmetry is the x axis, we have a sideways parabola,

[tex] {y}^{2} = x[/tex]

Since the vertex is at the orgin, our answer will be

[tex]a(y - k) {}^{2} + h = 0[/tex]

[tex]a {y}^{2} = x[/tex]

It passes through (3,2).

[tex]a(2) {}^{2} = 3[/tex]

[tex]a(4) = 3[/tex]

[tex]a = \frac{3}{4} [/tex]

So our function is

[tex] \frac{3}{4} {y}^{2} = x[/tex]

Answer:

[tex]x=\dfrac34y^2[/tex]

Step-by-step explanation:

A graph with x-axis symmetry means that for every (x, y) point there is also a (x, -y) point on the graph, i.e. the x-axis acts like a mirror.  Therefore, it is a sideways parabola.

The general form of this type of equation is [tex]x=ay^2+by+c[/tex]

Since we know that the curve has a vertex of (0, 0) then [tex]c=0[/tex]

Therefore, [tex]x=ay^2+by[/tex]

Given that (3, 2) is on the curve, then (3, -2) is also on the curve.

Substituting the given point (3, 2):

[tex]3=a(2)^2+b(2)[/tex]

[tex]3=4a+2b[/tex]

Substituting the given point (3, -2)

[tex]3=a(-2)^2+b(-2)[/tex]

[tex]3=4a-2b[/tex]

Adding the equations together to eliminate [tex]b[/tex]:

[tex]\implies 6=8a[/tex]

[tex]\implies 3=4a[/tex]

[tex]\implies a=\dfrac34[/tex]

Therefore,

[tex]x=\dfrac34y^2+by[/tex]

Again, substituting point (3, 2) means that b = 0

Therefore, the final equation is:

[tex]x=\dfrac34y^2[/tex]

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