A quadratic function $f$ is given by $f(x) = ax^{2} + bx + c$ where $a$ is not $0$ . Select ali the statements that must be true about the graph of $f$ . $\newline$ (A) The $y$ -intercept of the graph is at $(0, c)$ . $\newline$ (B) The graph has an $x$ -intercept at $(c, 0)$ . $\newline$ (C) When $a<0$ the graph opens downward. $\newline$ (D) The graph has two $x$ -intercepts. $\newline$ (E) If $b=0$ , then the vertex is on the $y$ -axis.

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Algebra 2

Write equations of cosine functions using properties

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Q. A quadratic function

f

is given by

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

where

a

is not

0

. Select ali the statements that must be true about the graph of

f

\newline

(A) The

y

-intercept of the graph is at

(0, c)

\newline

(B) The graph has an

x

-intercept at

(c, 0)

\newline

a<0

the graph opens downward.

\newline

(D) The graph has two

x

-intercepts.

\newline

(E) If

b=0

, then the vertex is on the

y

-axis.

Y-Intercept Definition: The y-intercept of a graph is the point where the graph crosses the y-axis. This occurs when $x = 0$ . Substituting $x = 0$ into the equation $f(x) = ax^2 + bx + c$ gives $f(0) = a(0)^2 + b(0) + c = c$ . Therefore, the y-intercept is at $(0, c)$ .
X-Intercepts and Statement B: The x-intercepts of a graph are the points where the graph crosses the x-axis. These occur when $f(x) = 0$ . Setting $f(x) = 0$ gives $ax^2 + bx + c = 0$ . The statement B suggests that there is an x-intercept at $(c, 0)$ , which would imply that $f(c) = 0$ . However, $f(c) = ac^2 + bc + c$ , which is not necessarily zero unless $a$ , $b$ , and $c$ satisfy a specific relationship. Therefore, statement B is not necessarily true.
Parabola Direction: When $a < 0$ , the parabola opens downward because the coefficient $a$ determines the direction of the opening. If $a$ is positive, the parabola opens upward, and if $a$ is negative, the parabola opens downward. This is a fundamental property of quadratic functions.
Quadratic Function X-Intercepts: The graph of a quadratic function can have zero, one, or two x-intercepts depending on the discriminant $b^2 - 4ac$ . If the discriminant is positive, there are two real and distinct x-intercepts. If it is zero, there is exactly one x-intercept (the vertex). If it is negative, there are no real x-intercepts. Therefore, statement D is not necessarily true.
Quadratic Function Vertex: If $b = 0$ , the quadratic function simplifies to $f(x) = ax^2 + c$ . The vertex of this parabola is at the point where the derivative of $f$ with respect to $x$ is zero. Taking the derivative gives $f'(x) = 2ax$ . Setting $f'(x) = 0$ gives $x = 0$ . Therefore, the vertex is at $x = 0$ , which is on the $y$ -axis.