Question

Which of the following systems of equations can be used to find the points of intersection of the graphs of a quadratic equation with zeros at 0 and 4 and a vertex at (2, -4) and a linear equation with a slope of -3 and y-intercept (0,8)?

a) Quadratic: \(y = -x^2 + 4x - 4\), Linear: \(y = -3x + 8\)

b) Quadratic: \(y = x^2 - 4x - 4\), Linear: \(y = 3x + 8\)

c) Quadratic: \(y = -x^2 + 4x - 4\), Linear: \(y = 3x + 8\)

d) Quadratic: \(y = x^2 - 4x - 4\), Linear: \(y = -3x + 8\)

Answer 1

To find the points of intersection of a quadratic equation and a linear equation, we can set the two equations equal to each other and solve for x.

Step-by-step explanation:

To find the points of intersection of a quadratic equation and a linear equation, we can set the two equations equal to each other and solve for x. Let's consider the options:

a) Quadratic: = −² + 4 − 4,

Linear: = −3 + 8

In this case, we can set the two equations equal to each other:

−² + 4 − 4 = −3 + 8

This simplifies to:

−² + 7 − 12 = 0

Using the quadratic formula, we find the solutions:

= 3, = 4

Therefore, option a can be used to find the points of intersection.

Answer 2

The correct system of equations based on the provided information is option a) Quadratic: y = -x^2 + 4x - 4, Linear: y = -3x + 8.

Step-by-step explanation:

The student is asking for the correct pair of equations that represent the points of intersection between a quadratic and a linear function based on given characteristics. The quadratic has zeros at 0 and 4 and a vertex at (2, -4), and the linear equation has a slope of -3 and a y-intercept of (0,8).

The standard form of a quadratic equation is y = ax^2 + bx + c. Since the zeros of the quadratic are 0 and 4, we can write the quadratic function in its factored form as y = a(x)(x - 4). Given the vertex at (2, -4), we can determine that the quadratic equation opens downwards, which means 'a' is negative, and the vertex form of the equation would be y = a(x - 2)^2 - 4. Plugging in the vertex coordinates, we find a = -1, thus the quadratic equation is y = -(x - 2)^2 - 4 = -x^2 + 4x - 4.

For the linear equation, since the slope is -3 and the y-intercept is 8, the equation in slope-intercept form is y = -3x + 8. Therefore, the correct systems of equations that can be used to find the points of intersection is option a): Quadratic: y = -x^2 + 4x - 4, Linear: y = -3x + 8.