Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used. Match each quadratic function with its respective graph. Graph Functions Graph of quadratic functions on a coordinate plane. Upward parabola vertex is at (2, 1) in quadrant 1. Left slope at (1, 2), (0, 5) enters quadrant 3. Right slope is at (3, 2), (4, 5) in quadrant 1. arrow Both Graph shows downward parabola plotted on a coordinate plane. The parabola has vertex at (minus 1, 4). The parabola has left slope at (minus 3, 0) and right slope at (1, 0). arrowBoth Graph shows upward parabola plotted on a coordinate plane. The parabola has vertex at (3, minus 4). The parabola has left slope at (1, 0) and right slope at (5, 0). arrowBoth

Drag the tiles to the correct boxes to complete the pairs Not all tiles will be used Match each quadratic function with its respective graph Graph Functions Gra class=
Drag the tiles to the correct boxes to complete the pairs Not all tiles will be used Match each quadratic function with its respective graph Graph Functions Gra class=

Respuesta :

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)

What is a Quadratic Function?

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.

How do we correctly match the graphs?

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;
https://brainly.com/question/1214333
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