# A diagram shows a parabola.It crosses the x-axis at x=-4 and x=6 and crosses the y axis at y=48.Findthe equation of te parabola

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### 1 Answer

You need to use the standard form of quadratic function, whose graph is a parabola, such that:

`f(x) = ax^2 + bx + c`

The problem provides the information that parabola crosses x axis at x = -4 and x = 6, hence, you need to remember that parabola crosses x axis at `y = f(x) = 0` , such that:

`f(-4) = 0 => 16a - 4b + c = 0`

`f(6) = 0 => 36a + 6b + c = 0`

The problem provides the information that parabola crosses y axis at y = 48, hence, you need to remember that parabola crosses y axis at x = 0, such that:

`f(0) = 48 => c = 48`

Replacing 48 for c in the above two equations, yields:

`{(16a - 4b = -48),(36a + 6b = -48):}`

Using elimination method, you may multiplicate by 3 the top equation and you may multiplicate by 2 the bottom equation, such that:

`{(48a - 12b = -144),(72a + 12b = -96):}`

`120a = -240 => a = -240/120 => a = -2`

Replacing -2 for a in bottom equation, yields:

`-72 + 6b = -48 => 6b = 72 - 48 => 6b = 24 => b = 4`

**Hence, evaluating the quadratic equation of parabola, under the given conditions, yields **`f(x) = -2x^2 + 4x + 48.`