Analyzing Quadratic Function and Finding Coordinates
To find the coordinates of A, B, and C, we need to analyze the given parabola and the quadratic function represented by the equation \( y = -(x - 7)(x + 1) \).
First, let's expand this equation to find the standard form of the quadratic function:
\[ y = -(x^2 + 1x - 7x - 7) \]
\[ y = -(x^2 - 6x - 7) \]
\[ y = -x^2 + 6x + 7 \]
Now we will find the x-intercepts (A and C) by setting y to 0 and solving for x:
\[ 0 = -x^2 + 6x + 7 \]
Now we can factor the quadratic or use the quadratic formula to solve it. Since \( -x^2 + 6x + 7 \) can be factored easily, we will factor it:
\[ 0 = -(x - 7)(x + 1) \]
Setting each factor equal to zero gives us the x-intercepts:
\[ x - 7 = 0 \implies x = 7 \]
\[ x + 1 = 0 \implies x = -1 \]
This gives us the coordinates of A and C:
- A = (-1, 0) since the x-intercept is where y is zero.
- C = (7, 0) for the same reason.
Now, we need to find the vertex of the parabola (point B). The vertex form of a parabola is \( y = a(x - h)^2 + k \), where (h, k) is the vertex of the parabola. Since the given parabola is facing downwards (due to the negative sign in front of \( x^2 \)), the vertex is a maximum point. The x-coordinate of the vertex (h) of a parabola in standard form \( y = ax^2 + bx + c \) can also be found using the formula \( h = -\frac{b}{2a} \):
Here, a = -1 and b = 6, so:
\[ h = -\frac{b}{2a} \]
\[ h = -\frac{6}{2(-1)} \]
\[ h = -\frac{6}{-2} \]
\[ h = 3 \]
We can substitute x = 3 back into the expanded equation to find the y-coordinate of the vertex:
\[ y = -x^2 + 6x + 7 \]
\[ y = -3^2 + 6(3) + 7 \]
\[ y = -9 + 18 + 7 \]
\[ y = -9 + 25 \]
\[ y = 16 \]
So the coordinates of B, the vertex, are (3, 16).
In conclusion, the coordinates of the points are:
- A = (-1, 0)
- B = (3, 16)
- C = (7, 0)
