To find the vertex and [tex]\( x \)[/tex]-intercepts of the graph of the quadratic equation [tex]\( y = x^2 - 6x - 7 \)[/tex], we follow these steps:
1. Finding the [tex]\( x \)[/tex]-intercepts:
The [tex]\( x \)[/tex]-intercepts of a graph are the points where the equation equals zero ([tex]\( y = 0 \)[/tex]). Thus, we solve the equation:
[tex]\[ x^2 - 6x - 7 = 0 \][/tex]
Solving this quadratic equation for [tex]\( x \)[/tex]:
[tex]\[ (x + 1)(x - 7) = 0 \][/tex]
This gives two solutions:
[tex]\[ x + 1 = 0 \Rightarrow x = -1 \][/tex]
and
[tex]\[ x - 7 = 0 \Rightarrow x = 7 \][/tex]
So, the [tex]\( x \)[/tex]-intercepts are:
[tex]\[ (-1, 0) \text{ and } (7, 0) \][/tex]
2. Finding the vertex:
The vertex of a quadratic equation [tex]\( y = ax^2 + bx + c \)[/tex] is given by the formula for the vertex [tex]\( x \)[/tex]-coordinate:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For our equation [tex]\( y = x^2 - 6x - 7 \)[/tex], we have [tex]\( a = 1 \)[/tex] and [tex]\( b = -6 \)[/tex]. Plugging these values into the formula:
[tex]\[ x = -\frac{-6}{2 \cdot 1} = \frac{6}{2} = 3 \][/tex]
Now, to find the [tex]\( y \)[/tex]-coordinate of the vertex, we substitute [tex]\( x = 3 \)[/tex] back into the equation:
[tex]\[ y = (3)^2 - 6(3) - 7 \][/tex]
[tex]\[ y = 9 - 18 - 7 \][/tex]
[tex]\[ y = -16 \][/tex]
So, the vertex is:
[tex]\[ (3, -16) \][/tex]
Based on these calculations:
- The [tex]\( x \)[/tex]-intercepts are [tex]\((-1, 0)\)[/tex] and [tex]\( (7, 0) \)[/tex].
- The vertex is [tex]\( (3, -16) \)[/tex].
Therefore:
- For the [tex]\( x \)[/tex]-intercepts, the correct answer is:
[tex]\[ A. \text{ } x\text{-intercepts: } (-1, 0), (7, 0) \][/tex]
- For the vertex, the correct answer is:
[tex]\[ F. \text{ } Vertex: (3, -16) \][/tex]
