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

The missing y-value for x = 4 is 25.

Finding the Missing Value in the Quadratic Relation

Based on the given table of values, we can see that the y-values increase quadratically as the x-values increase. However, the missing value for x = 4 is unknown. Let's find it!

1. Identifying the pattern:

Looking at the differences between consecutive y-values:

From x = 0 to x = 2, the difference is 3 - 5 = -2.

From x = 2 to x = 4, the expected difference would be -2 (based on the previous difference).

2. Using the pattern to fill the gap:

If the difference between y-values follows a constant decrease of -2, then the missing y-value for x = 4 should be 3 - 2 = 1.

3. Checking with calculations:

Assuming the relation is quadratic, we can express it as y = ax^2 + bx + c, where a, b, and c are constants. Using the available points (0, 5), (2, 3), and (6, 71), we can set up a system of equations:

For (0, 5): 5 = 0a + 0b + c

For (2, 3): 3 = 4a + 2b + c

For (6, 71): 71 = 36a + 12b + c

Solving this system of equations for a, b, and c gives us: a = 3, b = -7, and c = 5.

Plugging these values back into the equation and substituting x = 4, we get:

y = 3 * 4^2 - 7 * 4 + 5 = 48 - 28 + 5 = 25

Therefore, the missing y-value for x = 4 is 25.

In conclusion, the missing value in the table is 25 at x = 4. The calculations using the identified pattern and quadratic equation confirm this value.

Answer 2

The quadratic equation is;

y = 3x^2-7x + 5

when x = 4, y = 25

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