Question

Write the equation for the parabola that has x− intercepts (1+ $\sqrt{5}$ ,0) and (1− $\sqrt{5}$,0) and passes through the point (4,8).

Answer 1

Using the Factor Theorem, the equation of the parabola is given by:

$f(x) = -\frac{1}{2}(x^2 - 2x - 24)$

The Factor Theorem states that a polynomial function with roots $x_1, x_2, ..., x_n$ is given by:

$f(x) = a(x - x_1)(x - x_2)...(x - x_n)$

• In which a is the leading coefficient.

In this problem:

• There are two x-intercepts, hence it is a 2nd degree equation, which's graph is a parabola.
• The zeroes are the x-intercepts, hence $x_1 = 1 + \sqrt{5}, x_2 = 1 - \sqrt{5}$.

Applying the Theorem, we have that:

$f(x) = a(x - 1 - \sqrt{5})(x - 1 + \sqrt{5})$

$f(x) = a[(x - 1) - \sqrt{5}][(x - 1) + \sqrt{5}]$

$f(x) = a(x^2 - 2x + 1 - 25)$

$f(x) = a(x^2 - 2x - 24)$

It through the point (4,8), which means that when $x = 4, f(x) = 8$, and this is used to find a.

$8 = a(4^2 - 2(4) - 24)$

$-16a = 8$

$a = -\frac{8}{16}$

$a = -\frac{1}{2}$

Hence, the equation of the parabola is:

$f(x) = -\frac{1}{2}(x^2 - 2x - 24)$

A similar problem is given at https://brainly.com/question/24380382