Question

a) show that x^2-8x+20 can be written in the form (x-a)^2+a where a is an integer
b) hence explain how you know that x^2-8x+20 is always positive

Answer 1

Answer:

a)

$x^2-8x+20$

$x^2-8x+16-16+20$

Completing the square

$(x-4)^2+4$

$(x-a)^2+a$

$a=4$

b)

You can prove that $x^2-8x+20$ is always positive, stating;

$x^2-8x+20=(x-4)^2+4=(x-4)^2+2^2$

The sum of two squared numbers can't be negative. Hence is always positive.

Also, if we take the discriminant of $x^2-8x+20$

$\Delta= \left(-8\right)^2-4\cdot \:1\cdot \:20\\\Delta= -16$

Once the discriminant is negative, the quadratic has no real root. So, the function never touches the real axis and the function lies above or below the real axis. If we take x as 0, y = 20, therefore, the function lies above the real axis. It is always positive.

Answer 2

The equation in the form $(x-a)^2+a$ is $(x-4)^2+36$

Vertex form of an equation:

Given the quadratic function $x^2-8x+20$

We are to write this quadratic function in vertex form as shown:

Group the given expression

• $(x^2+8x)-20$

Complete the square method

$=(x^2+8x -(8/2)^2+(8/2)^2)+20\\=(x^2+8x-4^2+4^2)+20\\=(x-4)^2+16+20\\=(x-4)^2+36$

Hence the equation in the form $(x-a)^2+a$ is $(x-4)^2+36$

Learn more on vertex form here: https://brainly.com/question/525947