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​

The graph of the equation has a minimum.

When y = 0, the solutions are

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:

.

This polynomial has roots such that , given by the following formulas:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

It's vertex is the point

In which

Where

If a<0, the vertex is a maximum point, that is, the maximum value happens at , and it's value is .

y = x2 – 8x + 4

Quadratic equation with

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

Solutions when y = 0

When y = 0, the solutions are

Extreme value:

The vertex. So

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

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°