Question

quadratic expressions and equations review

1. the lowest or highest point on a parabola is the ______.

a. vertex
b. slope
c. axis of symmetry
d. y - axis

1. for $y = ax^2 + bx + c$, if $a > 0$, then the parabola opens ____ but if $a < 0$, then the parabola opens ____.
2. which equation represents the parabola?

graph of a parabola opening downward with vertex at the origin
a. $y = \frac{1}{2}x^2$
b. $y = -\frac{1}{2}x^2$
c. $y = 2x^2$
d. $y = -2x^2$

1. how many real solutions does the equation $4x^2 + 4x + 1 = 0$ have?

a. 0
b. 1
c. 2
d. more than 2

Explanation:

Question 1

The vertex of a parabola is the lowest (if opening up) or highest (if opening down) point. Slope is for lines, axis of symmetry is a line, and y - axis is a coordinate axis. So the correct answer is the vertex.

For the quadratic function \(y = ax^{2}+bx + c\), the coefficient \(a\) determines the direction of the parabola's opening. If \(a>0\), the parabola opens upward (like a "U" shape), and if \(a < 0\), it opens downward (like an "n" shape).

1. First, check the direction of the parabola: The given parabola opens downward, so the coefficient of \(x^{2}\) should be negative. This eliminates options a (\(y=\frac{1}{2}x^{2}\)) and c (\(y = 2x^{2}\)).
2. Then, check the "width" of the parabola. The general form of a parabola is \(y=ax^{2}\). For \(y=-\frac{1}{2}x^{2}\), when \(x = 1\), \(y=-\frac{1}{2}\); for \(y=-2x^{2}\), when \(x = 1\), \(y=-2\). Looking at the graph, when \(x = 1\), the \(y\) - value is closer to \(-\frac{1}{2}\) than to \(-2\). So the equation is \(y=-\frac{1}{2}x^{2}\).

Answer:

a. vertex

Question 2