Question

Consider the following quadratic equation. y = x2 – 8x + 4 Which of the following statements about the equation are true? The graph of the equation has a minimum. When y = 0, the solutions of the equation are a = 4 + 2V3 o When y = 0, the solutions of the equation are r x = 8 + 2V2. o The extreme value of the graph is at (4,-12). The extreme value of the graph is at (8,-4). U The graph of the equation has a maximum. Submit​

Answer 1

Answer:

The graph of the equation has a minimum.

When y = 0, the solutions are $4 \pm 2\sqrt{3}$

The extreme value of the graph is (4,-12).

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

$f(x) = ax^{2} + bx + c$

It's vertex is the point $(x_{v}, y_{v})$

In which

$x_{v} = -\frac{b}{2a}$

$y_{v} = -\frac{\Delta}{4a}$

Where

$\Delta = b^2-4ac$

If a<0, the vertex is a maximum point, that is, the maximum value happens at $x_{v}$, and it's value is $y_{v}$.

y = x2 – 8x + 4

Quadratic equation with $a = 1, b = -8, c = 4$

a is positive, so it's graph has a minimum.

Solutions when y = 0

$\Delta = b^2-4ac = 8^2 - 4(1)(4) = 64 - 16 = 48$

$x_{1} = \frac{-(-8) + \sqrt{48}}{2} = \frac{8 + 4\sqrt{3}}{2} = 4 + 2\sqrt{3}$

$x_{2} = \frac{-(8) - \sqrt{48}}{2} = \frac{8 - 4\sqrt{3}}{2} = 4 - 2\sqrt{3}$

When y = 0, the solutions are $4 \pm 2\sqrt{3}$

Extreme value:

The vertex. So

$x_{v} = -\frac{-8}{2} = 4$

$y_{v} = -\frac{48}{4} = -12$

The extreme value of the graph is (4,-12).