Question - Analyzing Quadratic Function and Finding Coordinates
Solution:
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)
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