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2 years ago

Write an equation for a parabola that has a vertex (1, 2) and passes through the point (3, -6).

Mathematics

1 answer:

Alex2 years ago

6 0

Answer:

y= -2x²+4x.

Step-by-step explanation:

the common equation of the parabola is y=ax²+bx+c, where a, b, c - numbers.

1) the coordinates of the vertex are:

$x_0=-\frac{b}{2a} \ and \ y_0=\frac{4ac-b^2}{4a}.$

2) if according to the condition x₀=1, then b= -2a.

If according to the condition y₀=2, then $2=\frac{4ac-b^2}{4a}=\frac{4ac-4a^2}{4a}=c-a;$

it means that c=a+2.

3) if b= -2a; c=a+2 and the point (3;-6) belongs to the given parabola, then it is possible to substitute them into the common equation of the parabola:

-6=3²*a-6a+a+2; ⇔ a= -2.

4) if a= -2, then b=-2a=4, and c=a+2=0.

5) if a=-2, b=4 and c=0, then the required equation of the parabola is:

y= -2x²+4x.

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