Question

I need help on my algebra project i have part A completed.

Answer 1

Solution

Given the function below

$f(x)=(x-a)(x-b)$

To create a new parabola in the pattern of the parabola above,

Let

$a=1\text{ and b}=1$

Thus, the above function becomes

$f(x)=(x-1)(x-1)$

Opening the brackets gives

$\begin{gathered} f(x)=x(x-1)-1(x-1) \\ f(x)=x^2-x-x+1 \\ f(x)=x^2-2x+1 \end{gathered}$

The leading coefficient is greater than 0 i.e

$x^2\text{ has a leading coefficient of 1 and it is greater than 0}$

Since the leading coefficient is greater than 0, the parabola opens upwards.

Hence, the direction of the parabola is upwards.

The zeros of the f(x) will be

$\begin{gathered} x-1=0 \\ x=1 \\ x-1=0 \\ x=1\text{ (twice)} \end{gathered}$

The zeros of the polymial is 1 (twice)

The graph of the created function is

From the graph above,

The y-intercept of the function is 1