Just as y = mx + b is a useful format for graphing linear functions, y = a(x - h)^2 + k is a useful format for graphing quadratic functions. We will explore its uses and learn how to convert any quadratic equation into vertex form.
A quadratic equation can be written as ax2 + bx + c = 0, as long as a does not equal 0. We usually see this form, known as standard form, because it is used for more than one method of solving quadratic equations. In this lesson, however, we will look at vertex form, which is written as a(x - h)2 + k. We can make good use of this format for graphing quadratic functions.
Vertex of a Quadratic Function
Let's graph the function associated with a quadratic equation: f'(x) = ax2 + bx + c or y = ax2 + bx + c.
Specifically, we can use y = 3x2 + 6x + 1 as an example.
Notice that the point at (-1,-2) is the lowest point on the graph. This is what we call the vertex. We can also find the coordinates of the vertex from the equation itself, using this formula: x = -b/2a.
In this case, x = -6/2(3) = -6/6 = -1. Hey, it works! Now since we know that x is -1, we can determine that y = 3(-1)2 + 6(-1) + 1 = 3(1) + (-6) + 1 = 3 - 6 + 1 = -3 + 1 = -2. So, there is our (-1,-2). This means that we do not have to graph the equation to find the vertex.
Converting to and from Vertex Form
Let's take the equation in standard form that we have already been working with:
y = 3x2 + 6x + 1
We already found the vertex to be (-1,-2).
The vertex form will look like:
a(x - h)2 + k
The a will be the same as in standard form, so we have:
y = 3(x - h)2 + k
Well, it turns out that the x-coordinate (our -1) is our h, and the y-coordinate (our -2) is the k. That gives us:
y = 3(x - (-1))2 + (-2), which is y = 3(x + 1)2 - 2
Thus, we use this method to convert from standard form to vertex form:
Find the x-coordinate using the formula x = -b/2a
Find the y-coordinate by evaluating f(x) = ax2 + bx + c with our value for x
Use the a from the standard form, the x-coordinate for h and the y-coordinate for k in y = a(x - h)2 + k
If we want to go the other way, we can simply multiply (x - h) by itself, multiply that result by a, and add k.
Now we have a vertex form. If we want to graph it, we can set things up in a systematic way:
Use the vertex as one of the points to plot on the graph
Try using a few x values that are less than the x-coordinate of the vertex and a few values that are greater than the x-coordinate of the vertex
Find the y-coordinate to go with each value of x
Plot the points, or enter them into a graphing program
Let's try it with a different example, like:
y= 2(x- 3)2 + 1
Since in this equation, 3 is h and 1 is k, the vertex is (3,1). Let's choose several values for x: 0, 1, 2, 3, 4, 5, 6.
If x is 0, then y = 2(0 - 3)2 + 1 = 2(-3)2 + 1 = 2(9) + 1 = 18 + 1 = 19
If x is 1, then y = 2(1 - 3)2 + 1 = 2(-2)2 + 1 = 2(4) + 1 = 8 + 1 = 9
If 'x is 2, then y = 2(2 - 3)2 + 1 = 2(-1)2 + 1 = 2(1) + 1 = 2 + 1 = 3
If x is 3, we already know that y is 1
If x is 4, then y = 2(4 - 3)2 + 1 = 2(1)2 + 1 = 2 + 1 = 3
Hey, that's just like the answer for x = 2. One less than the x-value of the vertex gives the same answer as one more than the x-value of the vertex.
If x is 5, then y = 2(5 - 3)2 + 1 = 2(2)2 + 1 = 2(4) + 1 = 8 + 1 = 9
See the pattern?
If x is 6, then y = 2(6 - 3)2 + 1 = 2(3)2 + 1 = 2(9) + 1 = 18 + 1 = 19
Here's how it looks when we plot the points:
The vertex form of a quadratic equation makes graphing relatively easy. If we are working with the graph or the vertex, this is the format to use. We can convert between vertex form and standard form with relative ease using the formula: x = -b/2a.
The vertex form of a quadratic function is an expression that easily provides the coordinates of the vertex point on the parabola.
The vertex form is
The vertex coordinates are
The vertex point is the extreme point on a parabola. If the quadratic term is positive, the parabola opens up, therefore the vertex is a minimum point.
And when the quadratic term is negative, the parabola opens down, and the vertex is a maximum point.
Quadratic functions often appear in optimization problems where we need to obtain the extreme value of the function or the coordinates of the vertex.
For example, to find the dimensions of a rectangle when a given perimeter and largest area are obtained by finding the x-coordinate of the vertex of a quadratic equation.
In general, the optimization problems simplify to finding a function whose extreme values are to be determined by imposing a given constraint.
To practice this technique, let's do the following problem.
Given a rectangular field that we want to fence with 200 yards, we will find the dimensions of the rectangular filed of the largest area if only three sides need to be fenced (the fourth side is along an existing wall).
To find the dimensions, perform the following steps.
1. Denoting the side of the field parallel to the wall with x and the other side with y, obtain a function of the area of the field in terms of x and y.
2. Provide an expression for the constraint.
3. Write the function obtained in 1. in terms of a single variable, using the constraint in 2.
4. Give the x- coordinate of the vertex corresponding to the quadratic function from 3.
1. A = xy
2. 2x + y = 200
3. A(x) = x(200 - 2x).
4. x = 50 yards, using a = -2, b = 200 to obtain the x-coordinate of the vertex.
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