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Write an equation for a parabola with x-intercepts at 0 and 1 and which passes through the point (2, -2).

I would greatly appreciate some help with this!

（っ＾▿＾）

Respuesta :

Answer:

y = -x(x - 1).

Step-by-step explanation:

y = a(x - b)(x - c) where a is a constant and b and c are the x-intercepts.

y = a(x - 0)( x - 1)

y =   ax(x - 1).

When x = 2, y = -2 so:

-2 = a(2) (2 - 1)

2a = -2

a = -1.

So the equation is y = -x(x - 1).

semsee45

Answer:

[tex]y=-x^2+x[/tex]

Step-by-step explanation:

Intercept form of a quadratic equation:

[tex]y=a(x-p)(x-q)[/tex]

where:

  • p and q are the x-intercepts.
  • a is some constant.

Given:

  • x-intercepts:  0 and 1
  • Point on the curve:  (2, -2)

Substitute the given values into the formula and solve for a:

[tex]\begin{aligned}y&=a(x-p)(x-q)\\\\\implies -2&=a(2-0)(2-1)\\-2&=a(2)(1)\\-2&=2a\\\dfrac{2a}{2}&=\dfrac{-2}{2}\\\implies a&=-1\end{aligned}[/tex]

Substitute the given x-intercepts and the found value of a into the formula:

[tex]y=-1(x-0)(x-1)[/tex]

Expand to standard form:

[tex]\implies y=-1(x)(x-1)[/tex]

[tex]\implies y=-x(x-1)[/tex]

[tex]\implies y=-x^2+x[/tex]

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