Question

What is the equation of the quadratic function with a vertex at (2.-25) and an x-intercept at (7.0)?

fix) = (x - 2)(x - 7)

fx) = (x + 2)(x + 7)

fx) = (x - 3)(x + 7)

fix) = (x+3)(x-7)

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Answer 1

`f(x) = (x+3)\cdot (x-7)`

Explanation:

According to the statement, the quadratic function (parabola) has the following form:

`y + 25 = C \cdot (x-2)^(2)`

The standard form is unleashed after expading the algebraic equation:

`y + 25 = C\cdot (x^(2)-4\cdot x +4)`

`y = C\cdot x^(2) - 4\cdot C \cdot x + (4\cdot C - 25)`

The zeroes of the second-order polynomial are contained in this expression:

`x = \frac{4\cdot C \pm \sqrt{16\cdot C^(2)- 4\cdot C \cdot (4\cdot C - 25)}}{2\cdot C}`

`x = 2 \pm \frac{5\cdot \sqrt {C}}{ C}`

Given that
`x = 7` and assuming a positive sign, the value of C is finally found:

`7\cdot C = 2\cdot C + 5\cdot √(C)`

`5\cdot C = 5 \cdot √(C)`

`C = +1` (since vertex is a minimum).

The remaining zero of the polynomial is:

`x = 2 - 5`

`x = -3`

Therefore, the polynomial is
`f(x) = (x+3)\cdot (x-7)`.