QUESTION IMAGE
Question
for each function below, choose the correct description of its graph. (a) choose the description for g(x) = -x - 2. vertical line horizontal line line with a negative slope line with a positive slope parabola opening downward parabola opening upward (b) choose the description for k(x) = 3x² - 4x. vertical line horizontal line line with a negative slope line with a positive slope parabola opening downward parabola opening upward (c) choose the description for f(x) = 4. vertical line horizontal line line with a negative slope line with a positive slope parabola opening downward parabola opening upward
Part (a)
Step1: Identify function type
The function \( g(x) = -x - 2 \) is a linear function in the form \( y = mx + b \), where \( m=-1 \) (slope) and \( b = -2 \) (y - intercept).
Step2: Analyze slope and line type
Since the slope \( m=-1<0 \), it is a line with a negative slope. It is not a vertical line (vertical lines have undefined slope and are of the form \( x = c \)) or a horizontal line (horizontal lines have slope 0 and are of the form \( y = c \)) or a parabola (parabolas are quadratic functions, degree 2, while this is degree 1).
Step1: Identify function type
The function \( k(x)=3x^{2}-4x \) is a quadratic function (degree 2) of the form \( y = ax^{2}+bx + c \), where \( a = 3\), \( b=-4\), \( c = 0 \). Quadratic functions graph as parabolas.
Step2: Determine parabola direction
For a quadratic function \( y=ax^{2}+bx + c \), if \( a>0 \), the parabola opens upward; if \( a < 0 \), it opens downward. Here, \( a = 3>0 \), so the parabola opens upward. It is not a line (linear functions are degree 1) or a vertical line (vertical lines are not quadratic).
Step1: Identify function type
The function \( f(x)=4 \) can be written as \( y = 4 \), which is a horizontal line (horizontal lines are of the form \( y = c \), where \( c \) is a constant).
Step2: Eliminate other options
It is not a vertical line (vertical lines are \( x = c \)), not a parabola (parabolas are quadratic, degree 2, this is degree 0) and since it is a horizontal line, it has slope 0, so it is not a line with negative or positive slope.
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line with a negative slope