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Vertex Form of a Quadratic Equation

Convert to vertex form of a quadratic equation by completing the square

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Vertex Form of a Quadratic Equation

Suppose a diver jumped into the ocean, and his path could be traced by the parabola \begin{align*}y=x^2-4x+2\end{align*}, with the value of \begin{align*}y\end{align*} representing the diver's distance above or below the surface of the water in feet. How far below the surface of the water would the diver descend? 

Vertex Form of a Quadratic Equation

There are several ways to write an equation for a parabola:

  • Standard form: \begin{align*}y=ax^2+bx+c\end{align*}
  • Factored form: \begin{align*}y=(x+m)(x+n)\end{align*}
  • Vertex form: \begin{align*}y=a(x-h)^2+k\end{align*}

So far, you have used both standard form and factored from. Now, we will use vertex form. The vertex form of a quadratic equation is \begin{align*}y=a(x-h)^2+k\end{align*}, where \begin{align*}(h,k)=\end{align*} vertex of the parabola and \begin{align*}a=\end{align*} leading coefficient

Let's determine the vertex of \begin{align*}y=-\frac{1}{2} (x-4)^2-7\end{align*} and state if it is a minimum or maximum:

Using the definition of vertex form, \begin{align*}h=4 \text{ and } k=-7\end{align*}.

  • The vertex is (4, –7).
  • Because \begin{align*}a\end{align*} is negative, the parabola opens downward.
  • Therefore, the vertex (4, –7) is a maximum point of the parabola.

Once you know the vertex, you can use symmetry to graph the parabola.

\begin{align*}x\end{align*} \begin{align*}y\end{align*}
2 -9
3 -7.5
4 –7
5 -7.5
6 -9

Now, let's write the equation for a parabola with \begin{align*}a=3\end{align*} and vertex (–4, 5) in vertex form:

Using the definition of vertex form, \begin{align*}y=a(x-h)^2+k, h=-4\end{align*} and \begin{align*}k=5\end{align*}.

\begin{align*}y &= 3(x-(-4))^2+5\\ y &= 3(x+4)^2+5\end{align*}

Finding the Vertex by Completing the Square

Consider the quadratic equation \begin{align*}y=x^2+4x-2\end{align*}. What is its vertex? You could graph this using your calculator and determine the vertex or you could complete the square.

Start by completing the square. Since \begin{align*}\frac{1}{2}b=\frac{1}{2}(4)=2, 2^2=4\end{align*}.

\begin{align*} && y&=x^2+4x-2\\ \text{Add 2 to each side} && y+2 &= x^2+4x\\ \text{Add 4 to each side} && y+2+4 &=x^2+4x+4 \\ \text{Factor the perfect square trinomial}&& y+6 &= (x+2)^2\\ \text{Subtract 6 from each side to get into vertex form}&& y&= (x+2)^2 -6 \end{align*}

The vertex is \begin{align*}(-2, -6)\end{align*}.

   

 

 

Examples

Example 1

Earlier, you were told that a diver who jumps into the ocean follows the path of the parabola \begin{align*}y=x^2-4x+2\end{align*}, with the value of \begin{align*}y\end{align*} representing the diver's distance above or below the surface of the water in feet. How far below the surface of the water would the diver descend?

To find how far below the surface the diver can go, we need to find the minimum for the parabola.

Start by completing the square. Since \begin{align*}\frac{1}{2}b=\frac{1}{2}(-4)=-2, (-2)^2=4\end{align*}.

\begin{align*} && y&=x^2-4x+2\\ \text{Subtract 2 from each side} && y-2 &= x^2-4x\\ \text{Add 4 to each side} && y-2+4 &=x^2-4x+4 \\ \text{Factor the perfect square trinomial}&& y+2 &= (x-2)^2\\ \text{Subtract 2 from each side to get into vertex form}&& y&= (x-2)^2 -2\end{align*}

The vertex is \begin{align*}(2, -2)\end{align*}.

The minimum value that the parabola reaches is the \begin{align*}y\end{align*}-value of the vertex, -2. Since we want to know how far below the surface the diver goes, we use the absolute value of the minimum value. The diver can go 2 feet below the surface of the water. 

Example 2

An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s. What is the maximum height that the arrow will reach and at what time will that happen?

The maximum height is the vertex of the parabola, when the parabola faces down. Therefore, we need to rewrite the equation in vertex form.

\begin{align*}\text{We rewrite the equation in vertex form.} && y &= -4.9t^2+50t+2\\ && y-2 &= -4.9t^2+50t\\ && y-2 &= -4.9(t^2-10.2t)\\ \text{Complete the square inside the parentheses.} && y-2-4.9(5.1)^2 &= -4.9(t^2-10.2t+(5.1)^2)\\ \text{Add 129.45 to get into vertex form.}&& y-129.45 &= -4.9(t-5.1)^2\\ && y &= -4.9(t-5.1)^2+129.45\end{align*}

Since the \begin{align*}y\end{align*}-value of the vertex is 129.45, then the maximum height is 129.45 meters.

Review

  1. Using the equation from the arrow in Example 2:
    1. How high will an arrow be four seconds after being shot? After eight seconds?
    2. At what time will the arrow hit the ground again?

Write the equation for the parabola with the given information.

  1. \begin{align*}a=a\end{align*}, vertex \begin{align*}=(h, k)\end{align*}
  2. \begin{align*}a=\frac{1}{3}\end{align*}, vertex \begin{align*}=(1, 1)\end{align*}
  3. \begin{align*}a=-2\end{align*}, vertex \begin{align*}=(-5, 0)\end{align*}
  4. Containing (5, 2) and vertex (1, –2)
  5. \begin{align*}a=1\end{align*}, vertex \begin{align*}=(-3, 6)\end{align*}

Rewrite each quadratic function in vertex form.

  1. \begin{align*}y=x^2-6x\end{align*}
  2. \begin{align*}y+1=-2x^2-x\end{align*}
  3. \begin{align*}y=9x^2+3x-10\end{align*}
  4. \begin{align*}y=32x^2+60x+10\end{align*}

For each parabola, find:

  1. The vertex
  2. \begin{align*}x-\end{align*}intercepts
  3. \begin{align*}y-\end{align*}intercept
  4. If it opens up or down
  5. The graph the parabola
  1. \begin{align*}y-4=x^2+8x\end{align*}
  2. \begin{align*}y=-4x^2+20x-24\end{align*}
  3. \begin{align*}y=3x^2+15x\end{align*}
  4. \begin{align*}y+6=-x^2+x\end{align*}
  5. \begin{align*}x^2-10x+25=9\end{align*}
  6. \begin{align*}x^2+18x+81=1\end{align*}
  7. \begin{align*}4x^2-12x+9=16\end{align*}
  8. \begin{align*}x^2+14x+49=3\end{align*}
  9. \begin{align*}4x^2-20x+25=9\end{align*}
  10. \begin{align*}x^2+8x+16=25\end{align*}
  11. Sam throws an egg straight down from a height of 25 feet. The initial velocity of the egg is 16 ft/sec. How long does it take the egg to reach the ground?
  12. Amanda and Dolvin leave their house at the same time. Amanda walks south and Dolvin bikes east. Half an hour later they are 5.5 miles away from each other and Dolvin has covered three miles more than the distance that Amanda covered. How far did Amanda walk and how far did Dolvin bike?
  13. Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.

Quick Quiz

  1. Graph \begin{align*}y=-3x^2-12x-13\end{align*}and identify:
    1. The vertex
    2. The axis of symmetry
    3. The domain and range
    4. The \begin{align*}y-\end{align*}intercept
    5. The \begin{align*}x-\end{align*}intercepts estimated to the nearest tenth
  1. Solve \begin{align*}y=x^2+9x+20\end{align*} by graphing.
  2. Solve for \begin{align*}x: 74=x^2-7\end{align*}.
  3. A baseball is thrown from an initial height of 5 feet with an initial velocity of 100 ft/sec.
    1. What is the maximum height of the ball?
    2. When will the ball reach the ground?
    3. When is the ball 90 feet in the air?
  1. Solve by completing the square: \begin{align*}v^2-20v+25=6\end{align*}

Review (Answers)

To see the Review answers, open this PDF file and look for section 10.6. 

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Vocabulary

vertex form of a quadratic function

The vertex form of a quadratic function is y=a(x-h)^2+k, where (h,k)= vertex of the parabola and a= leading coefficient.

Intercept

The intercepts of a curve are the locations where the curve intersects the x and y axes. An x intercept is a point at which the curve intersects the x-axis. A y intercept is a point at which the curve intersects the y-axis.

Parabola

A parabola is the characteristic shape of a quadratic function graph, resembling a "U".

Vertex

A vertex is a corner of a three-dimensional object. It is the point where three or more faces meet.

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