Question

a quadratic function is defined by $p(x) = (x - 1)(x + 3)$. using your graph from the previous question, identify the coordinates of the $x$-intercepts, the $y$-intercept, and the vertex. a. left-most $x$-intercept: $(-3, 0)$ b. right-most $x$-intercept: $(1, 0)$ c. $y$-intercept: $(0, -3)$ d. vertex: $(-1, -4)$

Explanation:

Step1: Find x-intercepts

To find the x-intercepts, set \( p(x) = 0 \). So, \( (x - 1)(x + 3)=0 \). This gives \( x - 1 = 0 \) or \( x + 3 = 0 \), so \( x = 1 \) or \( x=-3 \). The left - most x - intercept is \( (-3,0) \) (option A) and the right - most x - intercept is \( (1,0) \) (option B).

Step2: Find y-intercept

To find the y - intercept, set \( x = 0 \) in \( p(x)=(x - 1)(x + 3) \). Then \( p(0)=(0 - 1)(0 + 3)=(-1)\times3=-3 \). So the y - intercept is \( (0,-3) \) (option C).

Step3: Find the vertex

First, expand \( p(x)=(x - 1)(x + 3)=x^{2}+3x - x - 3=x^{2}+2x - 3 \). The x - coordinate of the vertex of a quadratic function \( y = ax^{2}+bx + c \) is given by \( x=-\frac{b}{2a} \). For \( p(x)=x^{2}+2x - 3 \), \( a = 1 \), \( b = 2 \), so \( x=-\frac{2}{2\times1}=-1 \). Substitute \( x=-1 \) into \( p(x) \): \( p(-1)=(-1)^{2}+2\times(-1)-3=1 - 2 - 3=-4 \). So the vertex is \( (-1,-4) \) (option D).

Answer:

A. left - most \( x \)-intercept: \( (-3,0) \)
B. right - most \( x \)-intercept: \( (1,0) \)
C. \( y \)-intercept: \( (0,-3) \)
D. vertex: \( (-1,-4) \)