Question

draw a line for the axis of symmetry of function f. also mark the x-intercept(s), y-intercept, and vertex of the function. f(x) = x² - 4x - 5

Explanation:

Step1: Find axis of symmetry

For quadratic \( f(x)=ax^2 + bx + c \), axis of symmetry is \( x = -\frac{b}{2a} \). Here, \( a = 1 \), \( b = -4 \), so \( x = -\frac{-4}{2\times1}=2 \).

Step2: Find x - intercepts

Set \( f(x)=0 \), so \( x^2 - 4x - 5 = 0 \). Factor: \( (x - 5)(x + 1)=0 \). Thus, \( x = 5 \) or \( x = -1 \).

Step3: Find y - intercept

Set \( x = 0 \), \( f(0)=0^2 - 4\times0 - 5=-5 \). So y - intercept is \( (0, -5) \).

Step4: Find vertex

Vertex is on axis of symmetry \( x = 2 \). Substitute \( x = 2 \) into \( f(x) \): \( f(2)=2^2 - 4\times2 - 5 = 4 - 8 - 5=-9 \). So vertex is \( (2, -9) \).

To draw:

• Axis of symmetry: vertical line \( x = 2 \).
• X - intercepts: mark \( (5, 0) \) and \( (-1, 0) \).
• Y - intercept: mark \( (0, -5) \).
• Vertex: mark \( (2, -9) \).

Answer:

Axis of symmetry: \( x = 2 \) (vertical line), x - intercepts \( (5, 0) \) and \( (-1, 0) \), y - intercept \( (0, -5) \), vertex \( (2, -9) \) (marked on the graph as per instructions).