Question

12. The table of values below describes a quadratic relation. Find the missing value and include calculations that demonstrate that this is the correct value.

Answer 1

Answer:

The quadratic equation is;

y = 3x^2-7x + 5

when x = 4, y = 25

Step-by-step explanation:

Mathematically, the general form of a quadratic equation is;

y = ax^2 + bx + c

From the first point, we have that x = 0

Thus;

5 = a(0)^2 + b(0) +c

c = 5

Now, in a case when x is 2

3 = a(2)^2 + b(2) + 5

3 = 4a + 2b + 5

4a + 2b = -2 ••••••(i)

when x = 6

71 = a(6)^2 + b(6) + 5

66 = 36a + 6b ••••••(ii)

From equation i,

2b = -2-4a ••••(iii)

we can rewrite equation ii as

66 = 36a + 3(2b)

Substitute iii into ii

66 = 36a + 3(-2-4a)

66 = 36a - 6 - 12a

66 + 6 = 36a -12a

24a = 72

a = 72/24

a = 3

Recall;

2b = -2-4a

2b = -2-4(3)

2b = -2-12

2b = -14

b = -14/2

b = -7

So the quadratic equation would be;

y = 3x^2 -7x + 5

Now let us test;

when x = 10, y = 235

That would be:

y = 3(10)^2 -7(10) + 5

y = 300 - 70 + 5

y = 235

This shows that the derived quadratic equation is correct

Now, when x = 4; we have it that;

y = 3(4)^2 -7(4) + 5

y = 48 -28 + 5

y = 20 + 5 = 25