IB MYP Year 4-5: Standard Mathematics : Unit 3: Function -Quadratic functions MYP Style Questions

Ans:

To show that the graphs of the functions $f(x) = 2x + 4$ and $g(x) = (x-1)^2 – 2$ intersect at points $A$ and $B$, we need to find the values of $x$ where these two functions are equal. Let’s set up the equation and solve it step by step.

We equate the two functions $f(x)$ and $g(x)$:

$2x + 4 = (x-1)^2 – 2$

Expanding the right side of the equation:

$2x + 4 = x^2 – 2x + 1 – 2$

Combining like terms:

$0 = x^2 – 4x – 5$

Now we have the equation $x^2 – 4x – 5 = 0$. This is a quadratic equation, and we can solve it by factoring, completing the square, or using the quadratic formula. Let’s solve it using the quadratic formula.

The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

For our equation $x^2 – 4x – 5 = 0$, we have $a = 1$, $b = -4$, and $c = -5$. Substituting these values into the quadratic formula, we get:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(-5)}}{2(1)}$

Simplifying further:

$x = \frac{4 \pm \sqrt{16 + 20}}{2}$

$x = \frac{4 \pm \sqrt{36}}{2}$

$x = \frac{4 \pm 6}{2}$

Now we have two possible solutions for $x$:

$x_1 = \frac{4 + 6}{2} = 5$

$x_2 = \frac{4 – 6}{2} = -1$

These are the $x$-coordinates of the points where the graphs of $f(x)$ and $g(x)$ intersect. Therefore, when the two graphs intersect, $x^2 – 4x – 5 = 0$ is satisfied.