Quadratic Function?

Question

The quadratic function (f) is defined by f(x) = x^2 + 8x +13

1)express f(x) in the form (x+p)^2 +q where p and q are integers

2)what is the least value for f(x) and the value of x for which this least value occurs

3)find an expression for f(x-3) in the form ax^2 +bx+c, where a,b and c are integers

Answer 1

PROBLEM 1:

This is called completing the square. Take the coefficient of the x term (8), divide it in 2 (4) and square it (16). Add this and subtract this from your first equation:

f(x) = x² + 8x + 13 + 16 - 16

Group the +16 with the x² and x terms:
f(x) = (x² + 8x + 16) + (13 - 16)

The first part becomes a perfect square:
f(x) = (x + 4)(x + 4) + (13 - 16)
f(x) = (x + 4)² + (13 - 16)

The second part is just a simple subtraction:
f(x) = (x + 4)² - 3

So there is your answer (p = 4, q = -3)

PROBLEM 2:

Looking at the equation above, (x+4)² will always be non-negative. It can't get smaller than 0 because it is a square. The smallest value will therefore occur when (x+4)² = 0

(x+4)² = 0
(x+4) = √0
x + 4 = 0
x = -4

Plugging it back into the equation:
f(-4) = ((-4) + 4)² - 3
f(-4) = (0)² - 3
f(-4) = 0 - 3
f(-4) = -3

So the answer is the minimum value occurs when:
x = -4
f(x) = -3

You can also confirm this by graphing. It will be a parabola centered around the vertical line x = -4 with its lowest point at (-4, -3). Also, in general, when you have an equation in the form:
f(x) = (x + p)² + q, the minimum point is (-p, q), so you can just "read" it from your equation.

PROBLEM 3:

f(x) = x² + 8x + 13
f(x-3) = (x-3)² + 8(x-3) + 13

First multiply out the squared term:
f(x-3) = (x² - 6x + 9) + 8(x-3) + 13

Then multiply out the 8(x-3) term:
f(x-3) = (x² - 6x + 9) + (8x -24) + 13

Now group like terms:
f(x-3) = x² + 8x - 6x + 9 - 24 + 13
f(x-3) = x² + 2x - 2