2.5 Quadratic Functions; Maxima and Minima
2.5.1 Learning Objectives
By the end of this section, you should be able to:
- identify the key features of a quadratic function;
- convert a quadratic function from standard form to vertex form;
- graph a quadratic using its vertex, axis of symmetry, and intercepts;
- calculate a quadratic function’s maximum or minimum value; and
- interpret an extreme value in context.
2.5.2 Forms of a Quadratic Function
A quadratic function has standard form
\[ f(x)=ax^2+bx+c, \]
where \(a\ne0\). Its graph is a parabola. In vertex form,
\[ f(x)=a(x-h)^2+k, \]
the vertex is \((h,k)\) and the axis of symmetry is \(x=h\). If \(a>0\), the parabola opens upward and has a minimum; if \(a<0\), it opens downward and has a maximum.
Example 2.23 Reading Vertex Form
Identify the vertex, axis of symmetry, opening direction, and extreme value of \(f(x)=-3(x+2)^2+7\).
Solution
Since \(x+2=x-(-2)\), the vertex is \((-2,7)\) and the axis is \(x=-2\). Because \(-3<0\), the parabola opens downward and has maximum value \(7\).
\[ \boxed{\text{Vertex }(-2,7),\quad \text{axis }x=-2,\quad \text{maximum }7} \]
Exercise 2.22 Identify the vertex, axis, opening direction, and extreme value of \(g(x)=2(x-4)^2-5\).
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Answer: The vertex is \((4,-5)\), the axis is \(x=4\), the parabola opens upward, and its minimum value is \(-5\).
2.5.3 Completing the Square
Completing the square converts standard form to vertex form. Factor the leading coefficient from the variable terms, add and subtract the square of half the \(x\)-coefficient inside the brackets, and simplify.
Example 2.24 Converting to Vertex Form
Write \(f(x)=2x^2-8x+3\) in vertex form.
Solution
\[ \begin{aligned} f(x)&=2(x^2-4x)+3\\ &=2(x^2-4x+4-4)+3\\ &=2\bigl((x-2)^2-4\bigr)+3\\ &=2(x-2)^2-5. \end{aligned} \]
\[ \boxed{f(x)=2(x-2)^2-5} \]
Exercise 2.23 Write \(f(x)=x^2+6x+1\) in vertex form.
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Answer: \(\boxed{f(x)=(x+3)^2-8}\)
2.5.4 Graphing Quadratic Functions
A useful sketch shows the vertex, axis of symmetry, intercepts, and opening direction. For \(f(x)=ax^2+bx+c\), the vertex has \(x\)-coordinate
\[ x=-\frac{b}{2a}. \]
Substitute this value into the function to find the vertex’s \(y\)-coordinate.
Example 2.25 Finding the Key Features
Find the vertex and intercepts of \(f(x)=x^2-4x-5\).
Solution
\[ x=-\frac{-4}{2(1)}=2,\qquad f(2)=-9, \]
so the vertex is \((2,-9)\). Also, \(f(0)=-5\), giving \(y\)-intercept \((0,-5)\). For the \(x\)-intercepts,
\[ x^2-4x-5=(x-5)(x+1)=0. \]
Thus, the \(x\)-intercepts are \((5,0)\) and \((-1,0)\). Since \(a>0\), the parabola opens upward.
\[ \boxed{\text{Vertex }(2,-9);\quad x\text{-intercepts }(-1,0),(5,0);\quad y\text{-intercept }(0,-5)} \]
Exercise 2.24 Find the vertex and intercepts of \(f(x)=-x^2+6x-8\).
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Answer: The vertex is \((3,1)\), the \(x\)-intercepts are \((2,0)\) and \((4,0)\), and the \(y\)-intercept is \((0,-8)\).
2.5.5 Maximum and Minimum Values
The \(y\)-coordinate of the vertex is the extreme value of a quadratic. In an application, check that the vertex lies in the meaningful domain and interpret both coordinates with units.
Example 2.26 Finding a Maximum Height
The height of a model rocket, in metres, \(t\) seconds after launch is \(h(t)=-5t^2+60t-40\). Find its maximum height.
Solution
\[ t=-\frac{60}{2(-5)}=6. \]
Then
\[ h(6)=-5(6)^2+60(6)-40=140. \]
\[ \boxed{\text{Maximum height }140\text{ m at }t=6\text{ s}} \]
Exercise 2.25 Find the maximum value of \(P(x)=-2x^2+24x+15\) and the value of \(x\) at which it occurs.
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Answer: The maximum occurs at \(x=6\), and the maximum value is \(\boxed{87}\).
2.5.6 Investigating Quadratics with Desmos
Desmos can verify algebraic work and estimate features. Enter the function, select its vertex and intercepts, and choose a window that shows the relevant graph. Graphical values are approximate unless supported by exact algebra.
Example 2.27 Estimating Features with Desmos
Graph \(f(x)=1.4x^2-3.2x-6\) and estimate its vertex and \(x\)-intercepts.
Solution
The vertex has \(x=8/7\approx1.14\), and substitution gives \(y\approx-7.83\). Desmos gives zeros of approximately \(-1.22\) and \(3.51\).
\[ \boxed{\text{Vertex }\approx(1.14,-7.83),\quad x\text{-intercepts }\approx(-1.22,0),(3.51,0)} \]
Exercise 2.26 Use Desmos to estimate the vertex and \(x\)-intercepts of \(f(x)=0.5x^2+2x-3\).
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Answer: The vertex is \((-2,-5)\), and the \(x\)-intercepts are approximately \((-5.16,0)\) and \((1.16,0)\).
2.5.7 Conceptual Takeaways
- Every quadratic graph has one vertex and one vertical axis of symmetry.
- Vertex form displays the extreme value and transformations directly.
- Standard form displays the \(y\)-intercept, while factored form displays real zeros.
- The leading coefficient determines the opening direction and whether the vertex is a maximum or minimum.
- A model’s domain and units are essential when interpreting an extreme value.
2.5.8 Skills You Should Be Able to Do
- Identify a quadratic’s vertex, axis, opening direction, intercepts, domain, and range.
- Convert standard form to vertex form by completing the square.
- Sketch a parabola from its key features.
- Calculate and interpret maximum or minimum values.
- Use Desmos to estimate and verify graph features.
2.5.9 Practice Problems with Solutions
Identify the vertex and axis of \(f(x)=3(x-2)^2+4\).
Show Solution
Comparing with \(a(x-h)^2+k\) gives
\[ \boxed{\text{vertex }(2,4),\quad \text{axis }x=2}. \]
State whether \(g(x)=-2(x+1)^2+9\) has a maximum or minimum, and give its value.
Show Solution
Since \(a=-2<0\), the parabola opens downward. Its vertex is \((-1,9)\), so
\[ \boxed{\text{maximum value }9}. \]
Write \(f(x)=x^2-10x+7\) in vertex form.
Show Solution
\[ f(x)=x^2-10x+25-25+7=(x-5)^2-18. \]
Thus, \(\boxed{f(x)=(x-5)^2-18}\).
Find the vertex of \(f(x)=2x^2+12x+11\).
Show Solution
\[ x=-\frac{12}{2(2)}=-3,\qquad f(-3)=-7. \]
Therefore, \(\boxed{(-3,-7)}\).
Find all intercepts of \(f(x)=x^2+x-12\).
Show Solution
The \(y\)-intercept is \((0,-12)\). Since
\[ x^2+x-12=(x+4)(x-3), \]
the \(x\)-intercepts are \((-4,0)\) and \((3,0)\).
\[ \boxed{x\text{-intercepts }(-4,0),(3,0);\quad y\text{-intercept }(0,-12)} \]
Determine the domain and range of \(f(x)=-4(x-3)^2+12\).
Show Solution
The function accepts every real input and has maximum value \(12\):
\[ \boxed{\text{domain }(-\infty,\infty),\quad \text{range }(-\infty,12]}. \]
The height of a ball is \(h(t)=-4.9t^2+19.6t+1.5\), where \(h\) is in metres and \(t\) in seconds. Find the maximum height.
Show Solution
\[ t=-\frac{19.6}{2(-4.9)}=2,\qquad h(2)=21.1. \]
Therefore, \(\boxed{21.1\text{ m at }2\text{ s}}\).
Find the zeros of \(f(x)=2x^2-3x-7\) to two decimal places.
Show Solution
\[ x=\frac{3\pm\sqrt{(-3)^2-4(2)(-7)}}{4} =\frac{3\pm\sqrt{65}}{4}. \]
Thus, \(\boxed{x\approx-1.27\text{ or }x\approx2.77}\).