Question

1. what is the vertex and y-intercept of the function f(x) = 1/2 x² - 5? options: x = 0, (-5, 0); x = -1, (0, 5); x = 1, (5, 0); x = 0, (0, -5)

Explanation:

Step1: Recall vertex form of parabola

The function \( f(x)=\frac{1}{2}x^{2}-5 \) is a quadratic function in the form \( f(x)=ax^{2}+bx + c \), where \( a=\frac{1}{2} \), \( b = 0 \), \( c=-5 \). The vertex of a parabola \( y=ax^{2}+bx + c \) has its x - coordinate given by \( x=-\frac{b}{2a} \). Since \( b = 0 \), \( x=-\frac{0}{2\times\frac{1}{2}}=0 \). To find the y - coordinate of the vertex, substitute \( x = 0 \) into the function: \( f(0)=\frac{1}{2}(0)^{2}-5=-5 \). So the vertex is \( (0,-5) \), but we are interested in the x - value of the vertex (the axis of symmetry) and the y - intercept.

Step2: Find the y - intercept

The y - intercept of a function is the value of the function when \( x = 0 \). Substitute \( x = 0 \) into \( f(x)=\frac{1}{2}x^{2}-5 \), we get \( f(0)=\frac{1}{2}(0)^{2}-5=-5 \). So the y - intercept is \( (0,-5) \) and the x - value of the vertex is \( x = 0 \).

Answer:

D. \( x = 0,(0,-5) \) (assuming the last option is D, as per the options given with \( x = 0,(0,-5) \))