A function is a rule or set of rules that assigns any input to a single output. We can write it as:
$$f(input)=output$$
$$f(word)=number\,of\,letters$$
$$burnt=5$$
The $f$ signifies a function. The right side is the function that is performed on the left side. If an operation more than 1 output then it is not a function. However, many inputs can result in one output. Mathematically, only if $a=b$ can $f(x)=a$ and $f(x)=b$.
The graph of a function has:
a horizontal axis; $x$ for input values
a vertical axis; $f(x)$ for output values
ordered pairs $(x, \, f(x))$ of input and output values
If a vertical line intersects a graph at more than one point then it is not a function.
The possible inputs for a function are called it’s domain. We can specify a domain with an interval. An interval is the set of numbers between 2 values. Eg: $0 \leq x \leq 1$. A function’s possible outputs are called it’s range. To indicate values that are included in an interval we use square bracket. For values to be excluded at the edges of the interval, we use parenthesis.
$$[2,4] \quad = \quad 2 \leq x \leq 4$$
$$(2,4) \quad = \quad 2<x<4$$
$$[0,∞) \quad = \quad x \geq 0$$
Since ∞ is unbounded, it cannot be included like a number. We use parenthesis when the value extends forever.
We use the union symbol ∪ as an ‘and’ if there are 2 ranges or domains:
$$(-∞,0) \, ∪ \, (0,∞)$$
Previously we said that we can make a parabola with $y=x^2$. Now we will expand with the following equation:$$f(x)=a(x-h)^2+k$$
This is a quadratic function. When graphed, it creates a parabola. This parabola is always perfectly symmetrical, with a fold line that goes right through the vertex, and intersects the $x$ axis exactly between the $x$ intercepts. A quadratic function is a ‘polynomial of the second degree’. A nomial is a term ($5x$, $6$, $3x^2$, etc), poly means many. So a polynomial is a collection of many terms, just like our equation above. ‘Second degree’ means the highest exponent (the highest power) is 2. At our equation above, the largest power is 2. Third degree would be a cubic function, something like $4x^3$, first degree would be a linear function like $4x$. Quadratics are found in the real world. It is used to work out the price of a good by a business; the graph of price against revenue forms a downward-opening parabola. The vertex represents the price at which the highest revenue is generated. Satellite dishes are a parabola shape, this is so that any radio wave that hits a part of the parabola head on is reflected into a focus point, thus concentrating weak signals onto a receiver. Car headlights work in the exact opposite. Anytime an object is thrown or shot or any kind of projectile fired/launched, the trajectory is a parabola and it’s path is dictated by a quadratic function.
The quadratic function above is known as the vertex form:
$a$ changes how narrow or wide the parabola is or if it is reflected in the $x$ axis. If $a>1$ then the parabola gets narrower. If $0<a<1$ then the parabola gets wider. If $a<0$ the parabola is reflected in the $x$ axis.
$h$ translates the parabola left or right. If $h>0$, the parabola shifts to the right. If $h<0$, parabola shifts to the left.
$k$ translates the parabola up or down. If $k>0$, parabola goes up. $k<0$, parabola goes down.
The vertex of the function is at $(h,k)$.
The quadratic function above can be written as:
$$f(x) = ax^2+bx+c$$
This second version is known as the standard form. The standard form is good for finding the $y$ intercept, as when $x=0$ then the $y$ intercept is $c$.
Lets look at this quadratic function:
$$f(x)=(x-3)^2-1$$
$a=1 \quad h=3 \quad k=-1$, the vertex is $(3,-1)$. It can be written in the standard form.
Step 1 is to multiply out the square: $x^2 -3x -3x +9$
Step 2 collect terms: $x^2 -6x+9$
Step 3 don’t forget $k$: $x^2 -6x+9-1$ goes to $x^2 -6x+8$
$$f(x)=x^2-6x+8$$
We already know the vertex from the previous vertex form, now we know that the graph crosses the $y$ axis at 8. To go from this standard for to vertex form we use a technique called Completing the Square.
Step 1. Ignore the constant and focus on the terms with $x$: $(x^2-6x)+8$
Step 2. Make a perfect square: Take the linear term’s number, half it and square it = 9
Step 3. Add this 9 inside the parenthesis: $(x^2-6x+9)+8$ But since we can’t just add something and not subtract (must be kept balanced):$f(x)=(x^2-6x+9)+8-9$.
Step 4. Find a number than when squared becomes 9 and when added together becomes -6. In our case this is $-3$. So, $(x^2-6x+9)=(x-3)^2$. $8-9=-1$. So $f(x)=(x-3)^2 -1$.
There is another form to this equation; the factored form:
$$f(x) = (x+d)(x+e)$$
It is good for showing us where the graph crosses the $x$ axis. To do this we need $f(x) = 0$. To get this we need either of those brackets to equal 0.
$(x+d) = 0$ when $d=-x$
$(x+e)=0$ when $e=-x$
The $x$ intercepts are $-e$ and $-d$.
To get from standard form into factored form we use a process called Factoring.
Step 1. Find 2 numbers that multiply to equal 8 and add upto -6. This is going to be $-2$ and $-4$. This is kind of directed trial and error. We can write out every pair that multiplies to 8. This will give us 4 options, but only 2 unique options. Then we can look at those number pairs and see what will add to give -6. In our case we negated the pairs.
Step 2. Write out: $f(x)=(x-2)(x-4)$
So by looking at this we can see that the graph will cross the $x$ axis at 4 and 2. As $2-2=0$ and $4-4+0$.
We were lucky with the above example that it is so easy to work on. Some can’t be factored. An expression in the form of $x^2 -a^2$ is known as a difference of 2 squares. It’s factorisation is $(x+a)(x-a)$. An expression in the form of $x^2 + a^2$ is a sum of 2 squares and can’t be factored with real numbers. A quadratic expression is a perfect square if it has the form $x^2 + 2ax + a^2$. It’s factorisation is $(x+a)^2$. If we are given a quadratic equation that can’t be factorised, we may be able to turn it into a perfect square.
$$x^2 +6x+4=0$$
This can’t be factored, but we can turn it into a perfect square by adding 5 to both sides:
$$x^2 +6x+9=5$$Now it is in the form$$x^2 + 2ax + a^2$$It factorises to: $$(x+3)^2 = 5$$Solve:$$\sqrt{(x+3)^2}=\sqrt{5}$$ $$x+3=\sqrt{5}$$ The $x$ intercepts:$$x=-3±\sqrt{5}$$
Another way to solve $x^2+bx+c=0$ is:
Step 1. Move $c$ to the other side: $x^2 +bx=-c$.
Step 2. Take the coefficient of $x$ which is $b$. Calculate $\frac{b}{2}$ then add $(\frac{b}{2})^2$ to both sides.
Step 3. Factor the perfect square.
Step 4. Solve for $x$.
We have found out how to find the $x$ intercept through the factored form and vertex form, but for more complicated quadratics, these methods won’t work. If the equation is in the form:$$ax^2 +bx + c=0$$We can solve for $x$ by completing the square:
Step 1. Move the constant $c$ over:$$ax^2+bx=-c$$Step 2. Divide all terms by $a$:$$x^2+\frac{b}{a}x=-\frac{c}{a}$$Step 3. Do what we did above.
Step 3.1. Coefficient of $x$ is $\frac{b}{a}$
Step 3.2. Divide by 2: $\frac{b}{2a}$
Step 3.3. Square it: $\frac{b^2}{4a^2}$
Step 3.4. Add that to both sides:
$$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$$
Step 4. The left side is now a perfect square and can be factored: $$(x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2}$$
Step 5. Clean up right side by combining terms. We need a common denominator; $4a^2$. We multiply $-\frac{c}{a}$ by $\frac{4a}{4a}$ (1): $$(x+\frac{b}{2a})^2 = \frac{-4ac}{4a^2}+\frac{b^2}{4a^2}$$Tidy further:$$(x+\frac{b}{2a})^2 = \frac{b^2-4ac}{4a^2}$$
Step 6. Square root both side:$$x+\frac{b}{2a} = ± \sqrt \frac{b^2-4ac}{4a^2}$$Step 7. Simplify the root:$$x+\frac{b}{2a} =± \frac{\sqrt{b^2-4ac}}{2a}$$Step 8. Isolate $x$: $$x =-\frac{b}{2a} ± \frac{\sqrt{b^2-4ac}}{2a}$$Step 9. Notice on the right there is a common denominator. Combine:$$x=\frac{-b± \sqrt{b^2-4ac}}{2a}$$This equation is known as the Quadratic Formula. It shows the $x$ intercepts. We can quickly see if there are any solutions. If the value inside the root is negative then there aren’t any real solutions. If the value inside the root is zero, there will be 1 solution. If the value is positive then there will be 2 solutions. In a quadratic formula, the value inside the root is called the discriminant.