Question

question 1 of 10
what are the vertex and x-intercepts of the graph of the function below?
y = x² - 6x - 7
a. vertex: (3, -2); intercepts: x = -1, 7
b. vertex: (3, -16); intercepts: x = -1, 7
c. vertex: (3, -16); intercepts: x = 1, -7
d. vertex: (3, -2); intercepts: x = 1, -7

Explanation:

Step1: Find the x - coordinate of the vertex

For a quadratic function \(y = ax^{2}+bx + c\), the x - coordinate of the vertex is given by \(x=-\frac{b}{2a}\). For the function \(y=x^{2}-6x - 7\), \(a = 1\), \(b=-6\), \(c=-7\).
So \(x=-\frac{-6}{2\times1}=\frac{6}{2}=3\).

Step2: Find the y - coordinate of the vertex

Substitute \(x = 3\) into the function \(y=x^{2}-6x - 7\).
\(y=(3)^{2}-6\times3 - 7=9-18 - 7=9-(18 + 7)=9 - 25=-16\). So the vertex is \((3,-16)\).

Step3: Find the x - intercepts

To find the x - intercepts, set \(y = 0\), so we solve the equation \(x^{2}-6x - 7=0\).
Factor the quadratic equation: \(x^{2}-6x - 7=(x - 7)(x+1)=0\).
Set each factor equal to zero: \(x - 7=0\) or \(x + 1=0\).
Solving \(x - 7=0\) gives \(x = 7\), and solving \(x + 1=0\) gives \(x=-1\). So the x - intercepts are \(x=-1\) and \(x = 7\).

Answer:

B. Vertex: \((3, - 16)\); Intercepts: \(x=-1,7\)