Question

practice with quadratic functions in standard form.
the graph of which function will have a maximum and a y-intercept of 4?

• ( f(x) = 4x^2 + 6x - 1 )
• ( f(x) = -4x^2 + 8x + 5 )
• ( f(x) = -x^2 + 2x + 4 )
• ( f(x) = x^2 + 4x - 4 )

Explanation:

Step1: Recall properties of quadratic functions

A quadratic function is in the form \( f(x) = ax^2 + bx + c \). The parabola opens up (has a minimum) if \( a>0 \) and opens down (has a maximum) if \( a<0 \). The \( y \)-intercept is the value of \( f(0) \), which is \( c \) (since \( f(0)=a(0)^2 + b(0)+c = c \)).

Step2: Check the coefficient \( a \) for maximum ( \( a<0 \)) and \( y \)-intercept \( c = 4 \)

• For \( f(x)=4x^2 + 6x - 1 \): \( a = 4>0 \) (opens up, has minimum), \( c=-1

eq4 \). Eliminate.

• For \( f(x)=-4x^2 + 8x + 5 \): \( a=-4<0 \) (opens down, has maximum), \( c = 5

eq4 \). Eliminate.

• For \( f(x)=-x^2 + 2x + 4 \): \( a=-1<0 \) (opens down, has maximum), \( c = 4 \). This satisfies both conditions.
• For \( f(x)=x^2 + 4x - 4 \): \( a = 1>0 \) (opens up, has minimum), \( c=-4

eq4 \). Eliminate.

Answer:

\( f(x)=-x^2 + 2x + 4 \) (the third option)