The correctly matched quadratic functions are given as follows:
- Part A) The function of the First graph is f(x) = (x-3) (x + 1)
- Part B) The function of the Second graph is f(x) = -2 (x - 1) ( x + 3)
- Part C) The function of the Third graph is f(x) = 0.5(x - 6) (x + 2)
<h3>What is a Quadratic Function?</h3>
Quadratics are polynomial equations of the second degree, which means that they contain at least one squared term.
Quadratic equations are another name for it. The quadratic equation has the following general form: ax2 + bx + c = 0.
<h3>How do we correctly match the graphs?</h3>
Recall that:
f(x) - a (x -c) (x - d)
where
- a is the leading coefficient
- c and d are the roots or zeros of the function.
Part A) First graph
We are given to know
The solutions or zeros of the first graph are
x=-1 and x=3
The parabola open up, so the leading coefficient a is positive, the function therefore, is equal to
- f(x) = (x-3) (x + 1); Find the value of the coefficient a.
The vertex is equal to the point (1, -4)
Substitute and solve for a
-4 = a(1-3)(1+1)
-4 = a(-2)(2)
a = 1
Hence, the function is equal to f(x) = (x-3) (x + 1)
Part B) Second Graph
We are also given to know that
The solutions or zeros of the first graph are
x=-3 and x=1
The parabola open down, so the leading coefficient a is negative
The function is equal to f(x) = a (x-1)(x+3)
Find the value of the coefficient a
The vertex is equal to the point (-1,8)
to solve for a, we must substitute:
8 = a(-1-1) (-1+3)
8 = a(-2)(2)
a = -2
Hence, the function is equal to: f(x) = -2 (x - 1) ( x + 3)
Part C)
We are given to know that the solutions or zeros of the first graph are
x=-2 and x=6
The parabola open up, so the leading coefficient a is positive
The function is equal to f(x) = a(x-6) (x+2).
Find the value of the coefficient a
The vertex is equal to the point (2,-8)
We substitute and solve for a:
-8 = a(2-6) (2+2)
-8 = a(-4)(4)
a = 0.5
Hence, the function is equal to f(x) = 0.5(x - 6) (x + 2)
Learn more about quadratic functions at;
brainly.com/question/1214333
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