Definition, domain and range, graphical representation
Quadratic function
A quadratic function is any function of the following form:
f(x)=ax^2+bx+c
a,b,c are all real numbers and a\not =0.
The following functions are quadratic:
- f(x)=x^2+2x+3
- g(x)=x^2
- h(x)=4x^2-2x
The graph of a quadratic function is a parabola.
If f(x)=ax^2+bx+c, then:
- The parabola faces up if a>0.
- The parabola faces down if a<0.
Vertex
The vertex of a parabola is the point where the parabola switches from increasing to decreasing (or from decreasing to increasing).
If f(x)=ax^2+bx+c is a quadratic function then the range of f(x) is determined by the vertex of the parabola.
Forms of a quadratic function
A quadratic function can be presented in three forms:
- standard form
- vertex form
- intercept form
Standard form of a quadratic function
Standard form of a quadratic function
The standard form of a quadratic function is:
f(x)=ax^2+bx+c
The following quadratic equation is in standard form:
f(x)=x^2-2x+1
Vertex form of a quadratic function
Vertex form of a quadratic function
The vertex form of a quadratic function is:
f\left(x\right)=a\left(x-h\right)^2+k
a,h,k are real numbers.
Observe the following:
f\left(x\right)=a\left(x-h\right)^2+k=a\left(x^2-2hx+h^2\right)+k=ax^2-\left(2ah\right)x+\left(ah^2+c\right)
It is indeed a quadratic function.
The vertex form of a quadratic function is very useful.
Consider the following function:
f(x)=a(x-h)^2+k
If it is in vertex form then the point (h,k) is the vertex of the parabola. It follows that the range of f(x) is either:
- [k,\infty) if a>0
- (-\infty,k] if a<0
Consider the following quadratic function:
f(x)=-3(x-2)^2+7
The vertex of the quadratic function is at (2,-7) and the range of f(x) is (-\infty, 7].
Consider the following function:
f(x)=2(x-4)^2+17
The quadratic function has a vertex at (4{,}17) and the range of f(x) is [17,\infty).
Suppose f(x)=ax^2+bx+c is a quadratic function in standard form. Then the vertex form of f(x) is:
f(x)=a\left(x+\dfrac{b}{2a}\right)^2+c-\dfrac{b^2}{4a}
In particular, the vertex of f(x) is at \left(\dfrac{-b}{2a}, c-\dfrac{b^2}{4a}\right)
- If a>0 then the range of f(x) is \left[c-\dfrac{b^2}{4a},\infty\right)
- If a<0 then the range of f(x) is \left(-\infty,c-\dfrac{b^2}{4a}\right]
Consider the following function:
Let f(x)=x^2-4x+3
Then f(x) is in standard form with :
- a=1
- b=-4
- c=3
The vertex form of f(x) is then computed as:
f\left(x\right)=a\left(x+\dfrac{b}{2a}\right)^2+c-\dfrac{b^2}{4a}\\=\left(x+\dfrac{-4}{2}\right)^2+3-\dfrac{16}{4}\\= \left(x-2\right)^2+3-4\\=\left(x-2\right)^2-1
Therefore the vertex of f(x) is at (2,-1) and the range of f(x) is [-1,\infty).
Intercept form of a quadratic function
x-Intercept form
The x-intercept form of a quadratic equation is:
f\left(x\right)=a\left(x-\alpha_1\right)\left(x-\alpha_2\right)
Where a,\alpha_1, \alpha_2 are real numbers.
The following quadratic function is in intercept form:
f(x)=(x-1)(x+1)
The following quadratic function is in intercept form:
f(x)=2(x-3)^2
The following quadratic function is in intercept form:
f(x)=-7(x-1)x
If f\left(x\right)=a\left(x-\alpha_1\right)\left(x-\alpha_2\right) is a quadratic function in x -intercept form, then the points \left(\alpha_1{,}0\right) and \left(\alpha_2{,}0\right) are the x -intercepts of the graph of f\left(x\right).
The quadratic function f(x)=2(x-1)(x+17) has x-intercepts at (1{,}0) and (-17{,}0).
Discriminant
Let f\left(x\right)=ax^2+bx+c be a quadratic function in standard form. The discriminant of f\left(x\right) is:
b^2-4ac
Consider the following quadratic function:
f\left(x\right)=2x^2-7x+1
The discriminant of f\left(x\right) is:
\left(-7\right)^2-4\left(2\right)\left(1\right)=49-8=41
Consider the following quadratic function:
f(x)=ax^2+bc+c
Let d=b^2-4ac be the discriminant of f\left(x\right). We then have the following:
- If d \gt 0 then f\left(x\right) has two x -intercepts.
- If d=0 then f\left(x\right) has one x -intercept.
- If d \lt 0 then f\left(x\right) does not have an x -intercept.
Furthermore, if d\geq 0 then the x -intercept form of f\left(x\right) is:
f(x)=ax^2+bx+c=a(x-\alpha_1) (x-\alpha_2)
Where:
- \alpha_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}
- \alpha_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}
The above formulas for finding the x-values of the x-intercept form of a quadratic equation is known as the quadratic formula.
Consider the following quadratic function:
f(x)=x^2-3x-10
To find its x-intercept form we factor the quadratic x^2-3x-10 as follows:
x^2-3x-10=(x-\alpha_1)(x-\alpha_2)
We have:
- \alpha_1=\dfrac{-(-3)+\sqrt{(-3)^2-4(1)(-10)}}{2(1)}=\dfrac{3+\sqrt{9+40}}{2}=\dfrac{3+7}{2}=5
- \alpha_2=\dfrac{-(-3)-\sqrt{(-3)^2-4(1)(-10)}}{2(1)}=\dfrac{3-7}{2}=-2
Therefore the quadratic function has x-intercept form that is:
f(x)=1(x-5)(x-(-2))=(x-5)(x+2)
Consider the following quadratic function:
f(x)=x^2+1
It does not have an x-intercept form. This can be seen in one of two ways.
First, the square of any real number is nonnegative and therefore x^2+1\geq 1 for all x, hence the graph of f(x) does not have an x-intercept.
Also, compute the discriminant:
d=0^2-4\left(1\right)\left(1\right)=-4
d\lt0, therefore the function has no x-intercept.
Suppose f\left(x\right)=x^2+bx+c has x -intercept form f\left(x\right)=\left(x-\alpha_1\right)\left(x-\alpha_2\right). Observe that:
\left(x-\alpha_1\right)\left(x-\alpha_2\right)=x^2-\left(\alpha_1+\alpha_2\right)x+\alpha_1\alpha_2
Therefore we have:
- \alpha_1+\alpha_2=-b
- \alpha_1\alpha_2=c
Thus when factoring a quadratic function, it is not necessary to use the quadratic formula. Instead we can try to find \alpha_1 and \alpha_2 satisfying the above equations.
Let f(x) be the following quadratic equation:
f(x)=x^2-4x+4
To factor this equation notice:
- (-2)\cdot (-2)=4
- (-2)+(-2)=-4
Therefore the quadratic equation factors as follows:
f(x)=(x-2)(x-2)=(x-2)^2
The graph of f(x) has one x-intercept at (2{,}0).
Suppose f(x)=x^2-c is a quadratic equation. Then f(x)=x^2+0b-c. Observe the following:
- \sqrt{c}+(-\sqrt{c})=0
- \sqrt{c}\cdot (-\sqrt{c})=-c
Therefore the quadratic equation has the following x-intercept form:
f(x)=x^2-c=\left(x-\sqrt{c}\right)\left(x+\sqrt{c}\right)
The quadratic equation f(x)=x^2-49 has x-intercept form:
f(x)=x^2-49=(x-7)(x+7)
Average rate of change and difference quotient
Average rate of change of a quadratic function
Let f(x)=ax^2+bx+c be a quadratic function. The average rate of change of f(x) from two real numbers x_0 to x_1 is :
\dfrac{f(x_1)-f(x_0)}{x_1-x_0}=a(x_0+x_1)+b
Let f(x)=x^2+7x+3. The average rate of change of f(x) from 3 to 5 is:
\dfrac{f\left(5\right)-f\left(3\right)}{5-3}\\=1\left(5+3\right)+7\\=8+7\\=15
Difference quotient of a quadratic function
Let f(x)=ax^2+bx+c be a quadratic function. The difference quotient of f(x) is:
\dfrac{f(x+h)-f(x)}{h}=2ax+b +ah
Consider the following function:
f(x)=x^2+7x+3
The difference quotient of f(x) is:
\dfrac{f\left(x+h\right)-f\left(x\right)}{h}\\=2\left(1\right)x+7+\left(1\right)h\\=2x+7+h
Quadratic equations and inequalities
Quadratic function
A quadratic equation is any equation equivalent to an equation of the following form:
ax^2+bx+c=0
The following is a quadratic equation:
2x^2+7x-3=0
Consider the following quadratic equation:
a(x-b)(x-c)=0
The solutions are:
- x=b
- x=c
Consider the quadratic equation:
(x-1)(x+3)=0
The solutions are:
- x=1
- x=-3
Consider the following equation:
x^2+7=3x+5
To solve it we begin by moving all terms to one side of the equation:
x^2+7-3x-5=0
x^2-3x+2=0
We then find the x-intercept form of the quadratic:
x^2-3x+2
By observing :
- (-2)\cdot (-1)=2
- (-2)+(-1)=-3
The quadratic equation is then solved as follows:
- x^2-3x+2=0
- (x-2)(x-1)=0
So the solutions to the equation are x=2 and x=1.
Not all quadratic equations have a solution.
Consider the following quadratic equation:
x^2+4=0
There is no solution because it does not have an x-intercept form.
Quadratic Inequality
A quadratic inequality is any inequality equivalent to an inequality of the following forms:
- ax^2+bx+c\geq 0
- ax^2+bx+c\gt 0
- ax^2+bx+c\leq 0
- ax^2+bx+c\lt 0
The following inequality is a quadratic inequality.
2x^2+3x-1\lt 0
Quadratic inequalities are solved by finding the x-intercept form of the quadratic.
Consider the following quadratic function:
f(x)=a(x-b)(x-c)
Suppose a>0 and b\leq c. The graph of the quadratic function has x-intercepts at (b,0) and (c,0).
- f(x) is only negative when b<x<c.
- f(x) is positive when x<b and when x>c.
Consider the following quadratic inequality:
2(x-1)(x+3)\leq 0
- When x=1 or x=-3, the quadratic is 0
- When x is between -3 and 1, the quadratic is less than 0
Therefore it has the following solution set:
-3\leq x \leq 1
Consider the following inequality:
x^2+7\gt 3x+5
It is solved as follows:
x^2+7\gt 3x+5
x^2+7-3x-5\gt 0
x^2-3x+2\gt 0
(x-1)(x-2)\gt 0
The solution set to the inequality is therefore:
(-\infty,1)\cup (2,\infty)
Consider the following quadratic inequality:
x^2+1\geq 0
The solution set to the quadratic inequality is all real numbers since the graph of f(x)=x^2+1 lies above the x-axis.
Consider the quadratic inequality:
x^2+1\lt 0
It has no solution because the function f(x)=x^2+1 is always positive.