Posted by guest on 27. August 2017 · Write a comment · Categories: math
A guest Post by Paul Pierce and Ashley Ross

With the advances in calculator technology, some developmental and college-level math courses are restricting the use of any type of graphing or programmable calculators. This is to help students avoid becoming dependent on their calculators for both simple arithmetic and graphing. So, some teachers are going “old school” and forbidding the use of calculators in the classroom. Therefore, it is imperative that students learn efficient methods for finding important values, as well as graphing functions, without the help of their calculator. One type of function that appears in many courses is the quadratic function, and one of the most critical points on the graph of a quadratic function is the vertex.

Fundamental Concepts of the Graph of a Quadratic Function

For the function $f(x)=ax^2+bx+c$ with $a\not=0$, the graph is a smooth, continuous curve called a parabola. This parabola opens upward if $a > 0$ or opens downward if $a < 0$. The vertex $(h,k)$ of the graph is the only turning point on the parabola, which makes it a critical point. The $y$-coordinate $k$ of the vertex represents the minimum value of the function if $a>0$, or the maximum value of the function if $a<0$. The point $(h,k)$ may be found using the formulas $h=\frac{-b}{2a}$ and $k=\frac{bh}{2}+c$, which begin to show the importance of the vertex. We give two examples:

Example 1. For $y=x^2+6x+3$, find the vertex $(h,k)$.

First find $h$ using $h=\frac{-b}{2a}=\frac{-6}{2(1)}=-3$.

Next find $k$ using $k=\frac{bh}{2}+c=\frac{(6)(-3)}{2}+3=-9+3=-6$.

So, the coordinates of the vertex of the parabola are $(-3, -6)$. Observe from the graph that this vertex is the lowest point on the parabola, which means that $k = -6$ is the minimum value of the function.

Example 2.For $y=-2x^2+8x-5$, find the vertex $(h,k)$.

First find h using $h=\frac{-b}{2a}=\frac{-8}{2(-2)}=2$.

Next find k using $k=\frac{bh}{2}+c=\frac{(8)(2)}{2}-5=8-5=3$.

So, the coordinates of the vertex of the parabola are $(2, 3)$. Note that this vertex is the highest point on the graph, which illustrates that $k = 3$ is the maximum value of this function.

Methods for Finding the Vertex of a Parabola

There are multiple ways of finding the coordinates of the vertex $(h,k)$. The easiest way to find $h$ is with the formula $h=\frac{-b}{2a}$, which is merely the first part of the quadratic formula. As for $k$, it is traditionally found by substituting $h$ into the function. Thus,
$$k=f(h)=ah^2+bh+c.\quad\quad(Eq\;1)$$
However, using $h=\frac{-b}{2a}$, the middle term, bh, can be rewritten as
$$bh=b\left({\frac{-b}{2a}}\right)=-\left({\frac{b^2}{2a}}\right)=-2a\left({\frac{b^2}{4a^2}}\right)=-2a\left({\frac{b}{2a}}\right)^2=-2ah^2.$$

So, $k=f(h)=ah^2+bh+c=ah^2-2ah^2+c=-ah^2+c$. Thus, a simpler formula for $k$ is
$$k=-ah^2+c.\quad\quad(Eq\;2)$$
Alternatively, Eq 1 can be written as $k=ah^2+bh+c=h(ah+b)+c$. Then, substituting $h=\frac{-b}{2a}$ into the expression $ah+b$ produces $ah+b=a\left({\frac{-b}{2a}}\right)+b=\frac{-b}{2}+b=\frac{b}{2}$. Replacing $ah+b$ with $\frac{b}{2}$ in the above equation gives $k=h(ah+b)+c=h\left({\frac{b}{2}}\right)+c=\frac{bh}{2}+c$. This “new” equation
$$k=\frac{bh}{2}+c\quad\quad(Eq\;3)$$
is the formula used in Example 1 and Example 2. It is possible to derive a formula for finding $k$ without first finding $h$. However, when finding the vertex, both $h$ and $k$ are required, so it is easier to remember Eq 1, Eq 2, or Eq 3.
Comparing Formulas for k:

  • Eq 1: $k=ah^2+bh+c$
  • Eq 2: $k=-ah^2+c$
  • Eq 3: $k=\frac{bh}{2}+c$

Additionally, Eq 3 can be re-written in various ways: $k=h\left({\frac{b}{2}}\right)+c$, or $k=b\left({\frac{h}{2}}\right)+c$, or $k=\frac{1}{2}bh+c$.

Examples

The following examples compare the different formulas for finding $k$.

Example 3. For $y=3x^2-24x-7$, find the vertex $(h,k)$.
First find $h=\frac{-b}{2a}=\frac{-(-24)}{2(3)}=4$ and compare formulas for k.

  1. $k=ah^2+bh+c$
    $k=3(4)^2-24(4)-7=3(16)-24(4)-7=48-96-7=-55$
  2. $k=-ah^2+c$
    $k=-3(4)^2-7=-3(16)-7=-48-7=-55$
  3. $k=\frac{bh}{2}+c$
    $k=\frac{(-24)(4)}{2}-7=(-24)(2)-7=-48-7=-55$

The vertex is $(h,k) = (4, -55)$. Since $a\ge0$, the parabola opens upward and this vertex represents the lowest point on the graph. Therefore, $k = -55$ is the absolute minimum value of this function.

Example 4. For $f(x)=-x^2-6x+1$, find the vertex $(h,k)$.
First find $h=\frac{-b}{2a}=\frac{-(-6)}{2(-1)}=-3$ and compare formulas for $k$.

  1. $k=ah^2+bh+c$
    $k=-1(-3)^2-6(-3)+1=-1(9)-6(-3)+1=-9+18+1=10$
  2. $k=-ah^2+c$
    $k=-(-1)(-3)^2+1=-(-1)(9)+1=9+1=10$
  3. $k=\frac{bh}{2}+c$
    $k=\frac{(-6)(-3)}{2}+1=(-3)(-3)+1=9+1=10$

The vertex is (h,k) = (-3, 10). Since $a\le 0$, the parabola opens downward and this vertex represents the highest point on the graph. Therefore, $k = 10$ is the absolute maximum value of this function.

Example 5.For $f(x)=3x^2-36x$, find the vertex $(h,k)$
First find $h=\frac{-b}{2a}=\frac{-(-36)}{2(3)}=6$ and compare formulas for $k$.

  1. $k=ah^2+bh+c$
    $k=3(6)^2-36(6)+0=3(36)-36(6)=108-216=-108$
  2. $k=-ah^2+c$
    $k=-3(6)^2+0=-3(36)=-108$
  3. $k=\frac{bh}{2}+c$
    $k=\frac{(-36)(6)}{2}+0=-36(3)=-108$

The vertex is (h,k) = (6, -108). Since $a\ge 0$, the parabola opens upward and this vertex represents the lowest point on the graph. Therefore, $k = -108$ is the absolute minimum value of this function.

Example 6. For $f(x)=5x^2+16x+3$, find the vertex $(h,k)$
First find $h=\frac{-b}{2a}=\frac{-16}{2\cdot5}=\frac{-8}{5}$ and compare formulas for k.

  1. $k=ah^2+bh+c$
    $k=5\left({\frac{-8}{5}}\right)^2+16\left({\frac{-8}{5}}\right)+3=5\left({\frac{64}{25}}\right)+16\left({\frac{-8}{5}}\right)+3\\=\frac{64}{5}-\frac{128}{5}+3=\frac{-64}{5}+3=\frac{-49}{5}$
  2. $k=-ah^2+c$
    $k=-5\left({\frac{-8}{5}}\right)^2+3=-5\left({\frac{64}{25}}\right)+3=\frac{-64}{5}+3=\frac{-49}{5}$
  3. $k=h\frac{b}{2}+c$, since h is a fraction.
    $k=\frac{-8}{5}\left({\frac{16}{2}}\right)+3=\frac{-64}{5}+3=\frac{-49}{5}$

The vertex is $(h,k)=\left({\frac{-8}{5},\frac{-49}{5}}\right)$. Since $a\ge 0$, the parabola opens upward the vertex represents the lowest point on the graph. Therefore, $k=\frac{-49}{5}$ is the absolute minimum value of the function.

Example 7.For $f(x)=29x^2+4x$, find the vertex $(h,k)$.
First find $h=\frac{-b}{2a}=\frac{-4}{2(28)}=\frac{-2}{29}$ and compare formulas for $k$.

  1. $k=ah^2+bh+c$
    $k=29\left({\frac{-2}{29}}\right)^2+4\left({\frac{-2}{29}}\right)+0=29\left({\frac{4}{841}}\right)-\frac{8}{29}=\frac{116}{84}-\frac{232}{84}=\frac{-116}{84}=\frac{-4}{29}$
  2. $k=-ah^2+c$
    $k=-29\left({\frac{-2}{29}}\right)^2+0=-29\left({\frac{4}{841}}\right)=\frac{-116}{84}=\frac{-4}{29}$
  3. $k=h\frac{b}{2}+c$, since h is a fraction
    $k=\frac{-2}{29}\left({\frac{4}{2}}\right)+0=\frac{-4}{29}$

The vertex is $(h,k)=\left({\frac{-2}{29},\frac{-4}{29}}\right)$. Since $a\ge 0$, the parabola opens upward and this vertex represents the lowest point on the graph. Therefore, $k=\frac{-4}{29}$ is the absolute minimum value of this function.

Example 8. For $f(x)=-41x^2+14x$, find the vertex $(h,k)$
First find $h=\frac{-b}{2a}=\frac{-14}{2(-41)}=\frac{7}{41}$ and compare formulas for $k$.

  1. $k=ah^2+bh+c$
    $k=-41\left({\frac{7}{41}}\right)^2+14\left({\frac{7}{41}}\right)=-41\left({\frac{49}{1681}}\right)+14\left({\frac{7}{41}}\right)\\=\frac{-2009}{1681}+\frac{4018}{1681}=\frac{2009}{1681}=\frac{49}{41}$
  2. $k=-ah^2+c$
    $k=41\left({\frac{7}{41}}\right)^2+0=41\left({\frac{49}{1681}}\right)=\frac{2009}{1681}=\frac{49}{41}$
  3. $k=h\frac{b}{2}+c$, since h is a fraction.
    $k=\frac{7}{41}\left({\frac{14}{2}}\right)+0=\frac{49}{41}$

The vertex is $(h,k)=\left({\frac{7}{49},\frac{49}{41}}\right)$. Since $a\le 0$, the parabola opens doenward and the vertex represents the highest point on the graph. Therefore, $k=\frac{49}{41}$ is the absolute maximum value of the function.

Example 9.For $h(t)=-16t^2+28t$, find the vertex $(h,k)$
First find $h=\frac{-b}{2a}=\frac{-128}{2(-16)}=4$ and compare formulas for $k$.

  1. $k=ah^2+bh+c$
    $k=-16(4)^2+128(4)+0=-16(16)+128(4)=-256+512=256$
  2. $k=-ah^2+c$
    $k=-(-16)(4)^2+0=(16)(16)=256$
  3. $k=\frac{bh}{2}+c$
    $k=\frac{(125)(4)}{2}+0=(128)(2)=256$

The vertex is $(h,k) = (4, 256)$. Since $a\le 0$, the parabola opens downward and this vertex represents the highest point on the graph. Therefore, $k = 256$ is the absolute maximum value of this function.

Example 10. Example #10: For $h(t)=-16t^2+120t$, find the vertex $(h,k)$
First find $h=\frac{-b}{2a}=\frac{-120}{2(-16)}=\frac{15}{4}$ and compare formulas for $k$.

  1. $k=ah^2+bh+c$
    $k=-16\left({\frac{15}{4}}\right)^2+120\left({\frac{15}{4}}\right)+0=-16\left({\frac{225}{16}}\right)+120\left({\frac{15}{4}}\right)\\=-225+30(15)=-225+450=225$
  2. $k=-ah^2+c$
    $k=-(-16)\left({\frac{15}{4}}\right)^2+0=16\left({\frac{225}{16}}\right)=225$
  3. $k=h\frac{b}{2}+c$
    $k=\left({\frac{15}{4}}\right)\left({\frac{120}{2}}\right)+0=(15)(15)=225$

The vertex is $(h,k)=\left({\frac{15}{4},225}\right)$. Since $a\le 0$, the parabola opens downward and this vertex represents the highest point on the graph. Therefore, $k = 225$ is the absolute maximum value of this function.

Implications

After comparing the three formulas for $k$, the only apparent advantage to using Eq 1 is that it is not a formula to remember, because it is just $f(h)$. However, it requires more work than the other two formulas. In particular, Eq 1 has 5 operations (1 square, 2 multiplications, and 2 additions), Eq 2 has 4 operations (1 square, 2 multiplications, and 1 addition), and Eq 3 has 3 operations (1 multiplication, 1 division, and 1 addition). Also, Eq 1 and Eq 2 require squaring $h$. For large values of $h$, this could be difficult without a calculator. If either $h$ or $b$ are even, then the fraction in Eq 3 reduces quickly, which simplifies the arithmetic. Additionally, for functions with a values greatly larger than $b$ values, using Eq 1 and Eq 2 are more difficult without a calculator compared to using Eq 3. The difference being Eq 1 and Eq 2 require squaring $h$ and Eq 3 does not. This is illustrated by Examples 7 and 8 above. So, although Eq 2 and Eq 3 are often computationally comparable, many times Eq 3 is more efficient.

Conclusion

If teachers of both developmental and college-level math want their students to be able to graph quadratic functions without relying on a calculator, then there needs to be an efficient method for finding the coordinates of the vertex. As the above examples illustrate, a very efficient method for finding the vertex $(h,k)$ is to use $h=\frac{-b}{2a}$ and $k=\frac{bh}{2}+c$. This new formula for finding k is easy to remember and it is easier to compute without a calculator than the traditional method of finding $f(h)$. The method is simple to teach and easy to learn, so it benefits both the teacher and the student.

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