Question

Which quadratic function is represented by the graph?

f(x) = 0.5(x + 3)(x − 1)
f(x) = 0.5(x − 3)(x + 1)
f(x) = 2(x + 3)(x − 1)
f(x) = 2 (x − 3)(x + 1)

Answer 1

`f(x) = 0.5(x+3)(x-1)`

Explanation:

GIVEN :-

A quadratic function is represented by the graph in which :-

• Vertex of the parabola = (-1 , -2)
• The function intersects x-axis at (-3 , 0) and (1 , 0)
• Y-intercept of the function = -1.5

TO FIND :-

• The quadratic function

GENERAL CONCEPT USED IN THIS QUESTION :-

A quadratic function has 2 forms :-

• General form → f(x) = ax² + bx + c
• Standard form → f(x) = a(x - h)² + k [∵ where h = x-coordinate of the vertex of the function & k = y-coordinate of the vertex of the function.]

SOLUTION :-

The quadratic function in the graph intersects x-axis at two points (-3 , 0) & (1 , 0). But there are infinite parabolas which also intersect the same two points. And those parabolas have their unique quadratic function.

Method 1 (System of equations method) -

To find the unique quadratic function , you need to use three points on the

curve so that you can form 3 equations & solve them.

Using the General form of quadratic function , substitute the known values for x & y.

Let the three points be -

• (-3 , 0)
• (1 , 0)
• (0 , -1.5)

Substitute (-3 , 0) in general form of function -

`0 = a(-3)^2 + b(-3) + c`

`=> 9a-3b+c = 0` (eqn.1)

Substitute (1 , 0) in general form of function -

`0 = a(1)^2 + b(1) + c`

`=> a + b +c =0` (eqn.2)

Substitute (0 , -1.5) in general form of function -

`-1.5 = a(0)^2 + b(0) + c`

`=> c = -1.5`

Substitute c = -1.5 in -

1) eqn.1 →
`9a - 3b - 1.5 = 0`

`=> 9a - 3b = 1.5`

`=> 3(3a-b) = 1.5`

`=> 3a-b = 0.5` (eqn.4)

2) eqn.2 →
`a+b-1.5=0`

`=> a+b=1.5` (eqn.5)

Add eqn.4 & eqn.5 to get the value of 'a'.

`(3a-b)+(a+b) = 0.5+1.5`

`=> 4a = 2`

`=> a = (2)/(4) = (1)/(2) = 0.5`

Substitute a = 0.5 in eqn.5 -

`0.5 + b = 1.5`

`=> b = 1.5 - 0.5 = 1`

Now, rewrite the function in general form by putting the values of 'a' , 'b' & 'c'.

`f(x) = (0.5)x^2 + (1)x - 1.5`

`=> f(x) = 0.5(x^2 + 2x - 3)`

Factorise the quadratic polynomial.

`=> f(x) = 0.5(x^2 + 3x - x - 3)`

`=> f(x) = 0.5[x(x+3)-1(x+3)]`

∴
`f(x) = 0.5(x+3)(x-1)`

Method 2 (Vertex method) -

Another way to find the function is by taking any point on the curve & using the vertex of the parabola ; substitute the known values for x , y , h & k in the Standard form of the function.

Let that point on the curve be (-3 , 0)

Vertex = (-1 , -2)

Substitute the values of x , y , h & k in Standard form of function.

`0 = a[-3 - (-1)]^2 + (-2)`

`=> 0 = 4a-2`

`=> 4a = 2`

`=> a = (2)/(4) = 0.5`

Now rewrite the Standard form of the function by putting the values of h , k & a.

`f(x) = 0.5(x+1)^2-2`

Expand it.

`=> f(x) = 0.5(x^2+2x+1)-2`

`=> f(x) = 0.5x^2+x+0.5-2`

`=>f(x) = 0.5x^2 + x- 1.5`

`=>f(x) = 0.5(x^2 + 2x - 3)`

Factorising it will give the final answer.

∴
`f(x) = 0.5(x+3)(x-1)`

Answer 2

• f(x) = 0.5(x + 3)(x − 1)

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hope it helps..

have a great day!!