A Quadratic Function is any function defined by a polynomial whose greatest exponent is two. That means it can be written in the form \(f(x)=ax^2+bx+c\), with the restrictions that the parameters \(a\), \(b\), and \(c\) are real numbers and \(a\) canNOT be zero. The graph of any quadratic function is a U-shaped curve called a parabola. There are certain key features that are important to recognize on a graph and to calculate from an equation.
Key features of a parabola
Example \(\PageIndex{1}\): Identify Features of a Parabola from a graph Determine features of th parabola illustrated below.
There are two important forms of a quadratic function
Definitions: Forms of Quadratic Functions A quadratic function is a polynomial function of degree two. The graph of a quadratic function is a parabola.
The graph of a quadratic in standard form, \(f(x)=a(x−h)^2+k\), is a graph of \(y=x^2\) that has been shifted horizontally \(h\) units and vertically \(k\) units. Thus the vertex, originally at at \((0,0)\), is located at the point \((h, k)\) in the graph of \(f\). A formula for the location of the vertex for a quadratic in general form can be found by equating the two forms for a quadratic. \[\begin{align*} a(x−h)^2+k &= ax^2+bx+c \\[4pt] ax^2−2ahx+(ah^2+k)&=ax^2+bx+c \end{align*} \] This is an identity, so it is true for ALL values of \(x\): the coefficients for the \(x^2\) terms must be the same, the coefficients for the \(x\) terms must be the same, and the constant terms must be the same. Equating the \(x\) coefficients we get \[–2ah=b \text{, so } h=−\dfrac{b}{2a}. \nonumber\] This is a formula for the \(x\)-coordinate of the vertex. The \(y\)-coordinate of the vertex is \(y = f(h)=k\). Features of the graph of a quadratic function depend on the parameter values \(a\), \(b\), \(c\) or \(a\), \(h\), \(k\) used in its equation. How features of the parabola for a quadratic function can be obtained is summarized below.
When the quadratic term, \(ax^2\), is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.
Example \(\PageIndex{2}\): Find the orientation of a parabola Determine whether each parabola opens upward or downward:
Determine whether the graph of each function is a parabola that opens upward or downward:
When given a quadratic in standard form \(f(x)=a(x−h)^2+k\), the vertex and axis of symmetry is easily found once the parameters \(h\) and \(k\) have been identified. (Notice the sign on \(h\)!!) The vertex is \((h, k)\) and the axis of symmetry is the vertical line \( x=h\). When given a quadratic in general form: \(f(x)=ax^2+bx+c\), more computation is required. After identifying parameters \(a\) and \(b\), calculate \(h=–\dfrac{b}{2a}\). Then find the corresponding \(y\) coordinate for that point on the graph: \( y=f(h)=k \). Once \(h\) and \(k\) have been determined, the vertex is at \((h, k)\) and the axis of symmetry is the vertical line \( x=h\).
Example \(\PageIndex{3a}\): Find the Vertex from the General Form of the Quadratic Equation For the graph of \(f(x)=3x^{2}-6 x+2\) find (a) the axis of symmetry and (b) the vertex. Solution: \( \begin{array}{llc} \text{a.} & \text{Identify the equation parameters} & a=3, b=-6, c=2 \\ & \text{The axis of symmetry is the vertical line } x=-\frac{b}{2 a} & \\ & \text{Substitute the values }a \text{ and } b \text{ into the formula} & x=-\frac{-6}{2 \cdot 3}=1 \\ && \text{The axis of symmetry is the line } x=1\\ \end{array} \) \( \begin{array}{llc} \text{b.} & \text{The vertex is a point on the line of symmetry, so} & \text{ The } x \text{ coordinate of the vertex is } x=1\\ & \text{The }y \text{ coordinate will be }f(1) & f(1)=3({\color{red}{1}})^2-6({\color{red}{1}})+2 \\ & \text{Simplify} & f(1) = 3-6+2 \\ & \text{The result is the }y\text{ coordinate of the vertex.} & f(1)=-1 \\ && \text{The vertex is } (1, -1)\\ \end{array} \)
Example \(\PageIndex{3b}\): Find the Vertex from the Standard Form of the Quadratic Equation For the graph of \(f(x)=6(x-3)^{2}+4\) find:
Solution: \( \begin{array}{lll} \text{a.} & \text{Identify the equation parameters} & a=6, h=3, k=4 \\ & \text{The axis of symmetry is the vertical line } x=h & \\ & \text{Substitute.} & \text{The axis of symmetry is the line } x=3\\ \\ \text{b.} &\text{Use the equation parameters} & a=6, h=3, k=4 \\ & \text{The vertex is the point } (h, k) & \text{The vertex is the point } (3,4)\\ \end{array} \)
For the following quadratic functions find a. the axis of symmetry and b. the vertex
Example \(\PageIndex{4}\) Find the minimum or maximum value of the quadratic function \(f(x)=x^{2}+2 x-8\). Solution: \( \begin{array}{llc} \text{Identify the equation parameters} & a=1, b=2, c=-8 \\ \text{Since }a \text{ is positive, the parabola opens upward. } & \text{The quadratic equation has a minimum.} \\ \text{State the formula for the axis of symmetry } & x=-\dfrac{b}{2 a} \\ \text{Substitute and simplify.} & x=-\dfrac{2}{2 \cdot 1}= -1 \quad \text{The axis of symmetry is the line } x=-1\\ & \qquad \qquad \qquad \qquad \text{ The } x \text{ coordinate of the vertex is } x=-1\\ \text{The }y \text{ coordinate will be }f(-1) & f(-1)=({\color{red}{-1}})^2-2({\color{red}{-1}})-8 \\ \text{Simplify} & f(-1) = 1-2-8 \\ \text{The result is the }y\text{ coordinate of the vertex.} & f(-1)=-9 \qquad \text{The vertex is } (-1, -9)\\ \end{array} \) Since the parabola has a minimum, the \(y\)-coordinate of the vertex is the minimum \(y\)-value of the quadratic equation. The minimum value of the quadratic is \(-9\) and it occurs when \(x=-1\).
Find the maximum or minimum value of the quadratic function
a. The minimum value of the quadratic function is \(−4\) and it occurs when \(x=4\). b. The maximum value of the quadratic function is \(5\) and it occurs when \(x=2\).
Any number can be the input value, \(x\), to a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all \(y\)-values greater than or equal to the \(y\)-coordinate at the vertex or less than or equal to the \(y\)-coordinate at the vertex, depending on whether the parabola opens up or down.
Example \(\PageIndex{5}\): Find the Domain and Range of a Quadratic Function Find the domain and range of \(f(x)=−5x^2+9x−1\). Solution As with any quadratic function, the domain is all real numbers. Because \(a\) is negative, the parabola opens downward and has a maximum value. \( h=−\dfrac{b}{2a} =−\dfrac{9}{2(-5)}=\dfrac{9}{10} \) The maximum value is given by \(f(h)\). \(f(\frac{9}{10})=−5(\frac{9}{10})^2+9(\frac{9}{10})-1 = \frac{61}{20}\) The range is \(f(x){\leq}\frac{61}{20}\), or \(\left(−\infty,\frac{61}{20}\right]\).
Find the domain and range of \(f(x)=2\Big(x−\frac{4}{7}\Big)^2+\frac{8}{11}\). AnswerThe domain is all real numbers. The range is \(f(x){\geq}\frac{8}{11}\), or \(\left[\frac{8}{11},\infty\right)\).
The \(y\)-intercept is the point where the graph crosses the \(y\) axis. All points on the \(y\)-axis have an \(x\) coordinate of zero, so the \(y\)-intercept of a quadratic is found by evaluating the function \(f(0)\). The \(x\)-intercepts are the points where the graph crosses the \(x\)-axis. All points on the \(x\)-axis have a \(y\) coordinate of zero, so the \(x\)-intercept of a quadratic can be found by solving the equation \(f(x)=0\). Notice in Figure \(\PageIndex{6}\) that the number of \(x\)-intercepts can vary depending upon the location of the graph.
Example \(\PageIndex{6}\): Finding the \(y\)- and \(x\)-intercepts of a General Form Quadratic Find the \(y\)- and \(x\)-intercepts of the quadratic \(f(x)=3x^2+5x−2\). Solution Find the \(y\)-intercept by evaluating \(f(0)\). \( f(0)=3(0)^2+5(0)−2 =−2 \) So the \(y\)-intercept is at \((0,−2)\). For the \(x\)-intercepts, find all solutions of \(f(x)=0\). \(0=3x^2+5x−2\) In this case, the quadratic can be factored easily, providing the simplest method for solution. Typically, quadratics in general form, like this example, are usually solved using factoring, or failing that, using the quadratic formula or complete the square. \(0=(3x−1)(x+2)\) \[\begin{align*} 0&=3x−1 & 0&=x+2 \\ x&= \frac{1}{3} &\text{or} \;\;\;\;\;\;\;\; x&=−2 \end{align*} \] So the \(x\)-intercepts are at \((\frac{1}{3},0)\) and \((−2,0)\).
Example \(\PageIndex{7}\): Find the \(y\)- and \(x\)-intercepts of the quadratic \( f(x)=x^2+x+2\). Solution Find the \(y\)-intercept by evaluating \(f(0)\). \( f(0)=(0)^2+(0)+2 =2 \) So the \(y\)-intercept is at \((0, \; 2)\). For the \(x\)-intercepts, find all solutions of \(f(x)=0\) or \( x^2+x+2 = 0\). Clearly this does not factor, so employ the quadratic formula. The quadratic formula: \(x=\dfrac{−b{\pm}\sqrt{b^2−4ac}}{2a}\) and for this equation, \(a=1\), \(b=1\), and \(c=2\). Substituting these values into the formula produces \[\begin{align*} x&=\dfrac{−1{\pm}\sqrt{1^2−4⋅1⋅(2)}}{2⋅1} \\ &=\dfrac{−1{\pm}\sqrt{1−8}}{2} = \dfrac{−1{\pm}\sqrt{−7}}{2} =\dfrac{−1{\pm}i\sqrt{7}}{2} \nonumber \end{align*}\] Since the solutions are imaginary, there are no \(x\)-intercepts.
Example \(\PageIndex{8}\): Find the \(y\)- and \(x\)-intercepts of a Standard Form Quadratic Find the \(y\)- and \(x\)-intercepts of the quadratic \(f(x)=-2(x+3)^2+5\). Solution Find the \(y\)-intercept by evaluating \(f(0)\). Notice that the quantity inside the parentheses (\((0+3)=(3)\)) is evaluated FIRST!!! \( \begin{align*} f(0) &=-2(0+3)^2+5\\ &=-2(3)^2+5\\ &=-2(9)+5\\ &=-18+5 = -13\\ \end{align*} \) So the \(y\)-intercept is at \((0,−13)\). For the \(x\)-intercepts, find all solutions of \(f(x)=0\). Solving a quadratic equation given in standard form, like in this example, is most efficiently accomplished by using the Square Root Property \( \begin{array}{c} 0 =-2(x+3)^2+5\\ 2(x+3)^2 =5\\ (x+3)^2 =\dfrac{5}{2}\\ x+3 =\pm \sqrt{ \dfrac{5}{2}}\\ x = -3 \pm \sqrt{ 2.5 }\\ \end{array} \) So the \(x\)-intercepts are at \( (-3+\sqrt{2.5}, 0) \) and \( (-3-\sqrt{2.5}, 0) \).
Find the \(y\)- and \(x\)-intercepts for the function \(g(x)=13+x^2−6x\). Answer\(y\)-intercept at \((0, 13)\), No \(x\)-intercepts
Details on how to find features of a quadratic function have been covered. Now, these features will be used to sketch a graph.
Example \(\PageIndex{9}\) How to Graph a General Form Quadratic Function Using Properties Graph \(f(x)=x^{2}-6x+8\) by using its properties. Solution:
Graph the following quadratic functions by using its properties.
Example \(\PageIndex{10}\): How to Graph a Vertex Form Quadratic Using Properties Graph the function \(f(x)=2(x+1)^{2}+3\) by using its properties Solution:
Graph the following functions using properties
As the above examples illustrate, it is often easier to graph a quadratic equation that is in standard (vertex) form, rather than in general form. This is particularly true when trying to find \(x\)-intercepts for equations that don't easily factor. There are two different approaches for transforming an equation in general form into an equation in standard (or vertex) form. One method uses the formulas for \(h\) and \(k\). The other method uses Complete the Square. Both will be illustrated below.
Example \(\PageIndex{11}\): Formula method of rewriting into standard form Rewrite the quadratic function \(f(x)=2x^2+4x−4\) into standard form. Solution Step 1. Values of the parameters in the general form are \(a=2\), \(b=4\), and \(c=-4\). Step 2. Solve for \(h\). \( h=−\dfrac{b}{2a} =−\dfrac{4}{2(2)} =−1 \) Step 3. Use the value found for \(h\) to find \(k\). \( k=f(h)=f(−1) =2(−1)^2+4(−1)−4 =−6 \) Step 4. The standard form of the function is: \[\begin{align*} f(x)&=a(x−h)^2+k \\ f(x)&=2(x+1)^2−6 \end{align*}\]
Find the standard form for the function \(g(x)=13+x^2−6x\). Answer\(g(x)=(x-3)^2 +4 \)
Another way of transforming \(f(x)=ax^{2}+bx+c\) into the form \(f(x)=a(x−h)^{2}+k\) is by completing the square. The latter form is known as the vertex form or standard form. This approach will also be used when circles are studied. We must be careful to both add and subtract the number to the SAME side of the function to complete the square. We cannot add the number to both sides as we did when we completed the square with quadratic equations. When we complete the square in a function with a coefficient of \(x^{2}\) that is not one, we have to factor that coefficient from the \(x\)-terms. We do not factor it from the constant term. It is often helpful to move the constant term a bit to the right to make it easier to focus only on the \(x\)-terms. Once we get the constant we want to complete the square, we must remember to multiply it by the coefficient that was part of the \(x^2\) term before we then subtract it.
Example \(\PageIndex{12}\): CTS method of rewriting into vertex form Rewrite \(f(x)=−3x^{2}−6x−1\) in the \(f(x)=a(x−h)^{2}+k\) form by completing the square. Solution:
Rewrite the following functions in the \(f(x)=a(x−h)^{2}+k\) form by completing the square.
So far we have started with a function and then found its graph. Now we are going to reverse the process. Starting with the graph, we will find the function.
Given a graph of a quadratic function, write the equation of the function in general form.
Example \(\PageIndex{13}\): Writing the Equation of a Quadratic Function from the Graph Write an equation for the quadratic function \(g\) in Figure \(\PageIndex{13}\) in standard (vertex) form and then rewrite the result into general form. Solution Since it is quadratic, we start with the \(g(x)=a(x−h)^{2}+k\) form. (Observe the minus sign in front of \(h\)!) The vertex, \((h,k)\), is \((−2,−3)\) so \(h=−2\) and \(k=−3\). Substituting theses values we obtain \(g(x)=a(x+2)^2–3\). Substituting the coordinates of a point on the curve, such as \((0,−1)\), we can solve for the parameter \(a\). \[\begin{align*} −1&=a(0+2)^2−3 \\ 2&=4a \\ a&=\dfrac{1}{2} \end{align*}\] Using the values found for parameters \(a,\) \(h,\) and \(k,\) write the standard form of the equation: \(g(x)=\dfrac{1}{2}(x+2)^2–3\). To write this in general polynomial form, we can expand the formula and simplify terms. \[\begin{align*} g(x)&=\dfrac{1}{2}(x+2)^2−3 \\ &=\dfrac{1}{2}(x+2)(x+2)−3 \\ &=\dfrac{1}{2}(x^2+4x+4)−3 \\ &=\dfrac{1}{2}x^2+2x+2−3 \\ g(x) &=\dfrac{1}{2}x^2+2x−1 \end{align*}\]
Example \(\PageIndex{14}\) Determine the quadratic function whose graph is shown. Solution: Since it is quadratic, we start with the \(f(x)=a(x−h)^{2}+k\) form. The vertex, \((h,k)\), is \((−2,−1)\) so \(h=−2\) and \(k=−1\). \(f(x)=a(x-(-2))^{2}-1 \qquad \longrightarrow \qquad f(x)=a(x+2)^{2}-1\) To find \(a\), we use the \(y\)-intercept, \((0,7)\). So \(x=0\) and \(f(0)=7\). \(7=a(0+2)^{2}-1\) Solve for \(a\). \(\begin{array}{l}{7=4 a-1} \\ {8=4 a} \\ {2=a}\end{array}\) Write the function. \(f(x)=2(x+2)^{2}-1\)
Write the quadratic function in the form \(f(x)=a(x−h)^{2}+k\) for each graph.
A coordinate grid has been superimposed over the quadratic path of a basketball in Figure \(\PageIndex{15}\). Find an equation for the path of the ball. Does the shooter make the basket? Figure \(\PageIndex{15}\): Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes. The path passes through the origin and has vertex at \((−4, 7)\), so \(h(x)=–\frac{7}{16}(x+4)^2+7\). To make the shot, \(h(−7.5)\) would need to be about 4 but \(h(–7.5){\approx}1.64\); so, he doesn’t make it.
axis of symmetry general form of a quadratic function standard form of a quadratic function vertex vertex form of a quadratic function zeros |