Question

question 3 of 25
what are the vertex and x-intercepts of the graph of the function given below?
y = x² - 2x - 48

a. vertex: (0, 0); intercepts: x = -4, 6
b. vertex: (-1, -21); intercepts: x = 6, 4
c. vertex: (1, -40); intercepts: x = 6, 7
d. vertex: (1, -49); intercepts: x = 8, -6

Explanation:

Step1: Find the vertex of the parabola

For a quadratic function \( y = ax^2 + bx + c \), the x - coordinate of the vertex is given by \( x=-\frac{b}{2a} \). In the function \( y=x^{2}-2x - 48 \), \( a = 1 \), \( b=-2 \), \( c=-48 \).
So, \( x=-\frac{-2}{2\times1}=\frac{2}{2} = 1 \).
To find the y - coordinate of the vertex, substitute \( x = 1 \) into the function:
\( y=(1)^{2}-2\times(1)-48=1 - 2-48=-49 \). So the vertex is \( (1,-49) \).

Step2: Find the x - intercepts

To find the x - intercepts, set \( y = 0 \), so we solve the equation \( x^{2}-2x - 48=0 \).
Factor the quadratic equation: \( x^{2}-2x - 48=(x - 8)(x + 6)=0 \) (since \( -8\times6=-48 \) and \( -8 + 6=-2 \)).
Set each factor equal to zero:
\( x-8 = 0\) gives \( x = 8 \); \( x + 6=0\) gives \( x=-6 \).

Answer:

D. Vertex: (1, -49); intercepts: \( x = 8, -6 \)