10.6: 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^24x+2\end{align*}
Guidance
There are several ways to write an equation for a parabola:
 Standard form: \begin{align*}y=ax^2+bx+c\end{align*}
y=ax2+bx+c  Factored form: \begin{align*}y=(x+m)(x+n)\end{align*}
y=(x+m)(x+n) 
Vertex form: \begin{align*}y=a(xh)^2+k\end{align*}
y=a(x−h)2+k
Vertex form of a quadratic equation: \begin{align*}y=a(xh)^2+k\end{align*}
Example A
Determine the vertex of \begin{align*}y=\frac{1}{2} (x4)^27\end{align*}
Solution:
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*}
a 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  
3  
4  –7 
5  
6 
Example B
Write the equation for a parabola with \begin{align*}a=3\end{align*}
Solution:
Using the definition of vertex form, \begin{align*}y=a(xh)^2+k, h=4\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+4x2\end{align*}
Example C
Find the vertex of \begin{align*}y=x^2+4x2\end{align*}
Solution:
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+4x2\\
\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*}
Guided Practice
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?
Solution:
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 vertext form.} && y &= 4.9t^2+50t+2\\
&& y2 &= 4.9t^2+50t\\
&& y2 &= 4.9(t^210.2t)\\
\text{Complete the square inside the parentheses.} && y24.9(5.1)^2 &= 4.9(t^210.2t+(5.1)^2)\\
\text{Add 129.45 to get into vertex form.}&& y129.45 &= 4.9(t5.1)^2\\
&& y &= 4.9(t5.1)^2+129.45\end{align*}
Since the \begin{align*}y\end{align*}
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK12 Basic Algebra: Solving Quadratic Equations by Completing the Square (14:06)
 Using the equation from the arrow in the Concept:
 How high will an arrow be four seconds after being shot? After eight seconds?
 At what time will the arrow hit the ground again?
Write the equation for the parabola with the given information.

\begin{align*}a=a\end{align*}
a=a , vertex \begin{align*}=(h, k)\end{align*}=(h,k) 
\begin{align*}a=\frac{1}{3}\end{align*}
a=13 , vertex \begin{align*}=(1, 1)\end{align*}=(1,1) 
\begin{align*}a=2\end{align*}
a=−2 , vertex \begin{align*}=(5, 0)\end{align*}=(−5,0)  Containing (5, 2) and vertex (1, –2)

\begin{align*}a=1\end{align*}
a=1 , vertex \begin{align*}=(3, 6)\end{align*}=(−3,6)
Rewrite each quadratic function in vertex form.

\begin{align*}y=x^26x\end{align*}
y=x2−6x 
\begin{align*}y+1=2x^2x\end{align*}
y+1=−2x2−x 
\begin{align*}y=9x^2+3x10\end{align*}
y=9x2+3x−10 
\begin{align*}y=32x^2+60x+10\end{align*}
y=32x2+60x+10
For each parabola, find:
 The vertex

\begin{align*}x\end{align*}
x− intercepts 
\begin{align*}y\end{align*}
y− intercept  If it opens up or down
 The graph the parabola

\begin{align*}y4=x^2+8x\end{align*}
y−4=x2+8x  \begin{align*}y=4x^2+20x24\end{align*}
 \begin{align*}y=3x^2+15x\end{align*}
 \begin{align*}y+6=x^2+x\end{align*}
 \begin{align*}x^210x+25=9\end{align*}
 \begin{align*}x^2+18x+81=1\end{align*}
 \begin{align*}4x^212x+9=16\end{align*}
 \begin{align*}x^2+14x+49=3\end{align*}
 \begin{align*}4x^220x+25=9\end{align*}
 \begin{align*}x^2+8x+16=25\end{align*}
 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?
 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?
 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
 Graph \begin{align*}y=3x^212x13\end{align*} and identify:
 The vertex
 The axis of symmetry
 The domain and range
 The \begin{align*}y\end{align*}intercept
 The \begin{align*}x\end{align*}intercepts estimated to the nearest tenth
 Solve \begin{align*}y=x^2+9x+20\end{align*} by graphing.
 Solve for \begin{align*}x: 74=x^27\end{align*}.
 A baseball is thrown from an initial height of 5 feet with an initial velocity of 100 ft/sec.
 What is the maximum height of the ball?
 When will the ball reach the ground?
 When is the ball 90 feet in the air?
 Solve by completing the square: \begin{align*}v^220v+25=6\end{align*}
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vertex form of a quadratic function
The vertex form of a quadratic function is , where vertex of the parabola and leading coefficient.Intercept
The intercepts of a curve are the locations where the curve intersects the and axes. An intercept is a point at which the curve intersects the axis. A intercept is a point at which the curve intersects the axis.Parabola
A parabola is the characteristic shape of a quadratic function graph, resembling a "U".Vertex
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Here you'll learn how to use the vertex form of a quadratic equation in order to find the vertex by completing the square.