Question

a quadratic function ( f(x) ) is hidden from view. you must find the ( y )-intercept(s) of ( f(x) ) and write the answer(s) in the form ( (x, y) ). choose the form of the quadratic function ( f(x) ) that you would like to see in order to answer the question most efficiently.
form: select a form
answer attempt 1 out of 2
there are no ( y )-intercepts

Explanation:

To find the \( y \)-intercept of a quadratic function \( f(x) \), we use the standard form of a quadratic function, which is \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a
eq 0 \).

Step 1: Recall the definition of the \( y \)-intercept

The \( y \)-intercept of a function is the point where the graph of the function intersects the \( y \)-axis. By definition, the \( x \)-coordinate of any point on the \( y \)-axis is \( 0 \). Thus, to find the \( y \)-intercept, we evaluate the function at \( x = 0 \).

Step 2: Evaluate \( f(x) = ax^2 + bx + c \) at \( x = 0 \)

Substitute \( x = 0 \) into the standard form:

$$ f(0) = a(0)^2 + b(0) + c $$

Simplify the expression:

$$ f(0) = 0 + 0 + c = c $$

Step 3: Interpret the result

The value \( f(0) = c \) means the \( y \)-intercept occurs at \( x = 0 \) and \( y = c \). Thus, the \( y \)-intercept of the quadratic function \( f(x) = ax^2 + bx + c \) is \( (0, c) \).

Why the standard form is efficient

The standard form \( f(x) = ax^2 + bx + c \) directly reveals the \( y \)-intercept (since the constant term \( c \) is the \( y \)-value when \( x = 0 \)). Other forms (e.g., vertex form \( f(x) = a(x - h)^2 + k \) or factored form \( f(x) = a(x - r_1)(x - r_2) \)) require additional steps to find the \( y \)-intercept (e.g., expanding the form or substituting \( x = 0 \)), making the standard form the most efficient choice.

Final Answer

To find the \( y \)-intercept, select the standard form \( \boldsymbol{f(x) = ax^2 + bx + c} \). The \( y \)-intercept is \( \boldsymbol{(0, c)} \) (where \( c \) is the constant term in the standard form). If the quadratic function is given (e.g., \( f(x) = 2x^2 + 3x + 5 \)), the \( y \)-intercept would be \( (0, 5) \).

Answer:

To find the \( y \)-intercept of a quadratic function \( f(x) \), we use the standard form of a quadratic function, which is \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a
eq 0 \).

Step 1: Recall the definition of the \( y \)-intercept

The \( y \)-intercept of a function is the point where the graph of the function intersects the \( y \)-axis. By definition, the \( x \)-coordinate of any point on the \( y \)-axis is \( 0 \). Thus, to find the \( y \)-intercept, we evaluate the function at \( x = 0 \).

Step 2: Evaluate \( f(x) = ax^2 + bx + c \) at \( x = 0 \)

Substitute \( x = 0 \) into the standard form:

$$ f(0) = a(0)^2 + b(0) + c $$

Simplify the expression:

$$ f(0) = 0 + 0 + c = c $$

Step 3: Interpret the result

The value \( f(0) = c \) means the \( y \)-intercept occurs at \( x = 0 \) and \( y = c \). Thus, the \( y \)-intercept of the quadratic function \( f(x) = ax^2 + bx + c \) is \( (0, c) \).

Why the standard form is efficient

The standard form \( f(x) = ax^2 + bx + c \) directly reveals the \( y \)-intercept (since the constant term \( c \) is the \( y \)-value when \( x = 0 \)). Other forms (e.g., vertex form \( f(x) = a(x - h)^2 + k \) or factored form \( f(x) = a(x - r_1)(x - r_2) \)) require additional steps to find the \( y \)-intercept (e.g., expanding the form or substituting \( x = 0 \)), making the standard form the most efficient choice.

Final Answer

To find the \( y \)-intercept, select the standard form \( \boldsymbol{f(x) = ax^2 + bx + c} \). The \( y \)-intercept is \( \boldsymbol{(0, c)} \) (where \( c \) is the constant term in the standard form). If the quadratic function is given (e.g., \( f(x) = 2x^2 + 3x + 5 \)), the \( y \)-intercept would be \( (0, 5) \).