WRITTEN LESSONS
STANDARD FORM-
The graph of a quadratic function is called a parabola. These parabolas can be graphed differently depending on the form that is being used. The parent function of quadratics is y = x^2. The first form that will be taught is the standard form. Standard form is a quadratic function that can be written in the standard form y = ax^2 + bx + c where a cannot equal zero. There are also parts of the parabola that need to be known. The axis of symmetry divides the parabola into mirror images and passes through the vertex, meaning you can find the vertex there. The lowest or highest point on a parabola is the vertex. The lowest is called the minimum and the highest is called the maximum. The parabola should look like a "U" shape. Back to the equation. Standard form consists of different parts of the equation need to be explained for a graph to be made. If "a" is positive then the parabola will open up. If "a" is negative, the parabola will open down. If a>1, then the parabola will be narrower than the parent function. If 0<a<1, then the parabola will be wider than the parent function. The axis of symmetry (x-value of vertex) can be found by using this formula x = -b/(2a). The y-intercept will equal "c" in the standard form formula. Instructions on how to graph standard form- Step 0: Label the "a", "b", and "c" values of the equation Step 1: Does it open up or down? (Depends on the "a" value) Step 2: Find the vertex: x value= -b/(2a), y value: Plug in x into standard form equation and solve. Step 3: Identify y-intercept and plot. Step 4: Sketch. |
Examples:
Example 1:
y= 3x^2+6x+2 a = 3 b = 6 c = 2 Step 1: Does it open up or down? It opens up because the "a" in the equation is positive. Step 2: Find the vertex: x= -b/(2a). x= -(6)/(2[3]) x= -1. This means that the axis of symmetry will be negative one, meaning the vertex will lie on the vertical line of negative one. Then plug the x into the equation to find the y value of the vertex. y= 3(-1)^2+6(-1)+2. y= 3-6+2. y= -1. This would make the vertex's ordered pairs (-1,-1) Step 3: Identify y-intercept and plot. The y-intercept is the "c"; therefore, the y-intercept is (0,2). Step 4: Sketch. Place the y-intercept point on the other side of the axis of symmetry with the same amount of units from the axis of symmetry. This must have the same units of space between each point for it to be a parabola and for it to be symmetrical. Then connect the points to have a "U" shape. Example 2:
y= -x^2+2x+3 a = -1 b = 2 c = 3 Step 1: Does it open up or down? It opens down because the "a" in the equation is negative. Step 2: Find the vertex: x= -b/(2a). x= -(2)/(2[-1]) x= 1. This means that the axis of symmetry will be positive one. Then plug the x into the equation to find the y value of the vertex. y= -(1)^2+2(1)+3. y= -1+2+3. y= 4. This would make the vertex's ordered pairs (1,4). Step 3: Identify y-intercept and plot. The y-intercept is the "c"; therefore, the y-intercept is (0,3). Step 4: Sketch. Place the y-intercept point on the other side of the axis of symmetry. This must have the same amount of space between each point for it to be a parabola and for it to be symmetrical. Then connect the points to have a "U" shape. Example 3: Word Problem
A swing is being built for a few toddlers in the schoolyard. The engineers created an outline of this using a quadratic equation y= 1/2x^2+2X-5/2 to create the perfect comfy swing. Create a graph that shows this parabola and then identify the lowest point of the parabola. a = 1/2 b = 2 c = -5/2 Step 1: Does it open up or down? It opens up because the "a" in the equation is positive. Even if it is a fraction, which would cause it to be wider than the parent function. Step 2: Find the vertex: x= -b/(2a). x= -(2)/(2[1/2]) x= -2. This means that the axis of symmetry will be negative two. Then plug the x into the equation to find the y value of the vertex. y= 1/2(-2)^2+2(-2)-5/2. y= 2-4-5/2. y= -9/2 or -4.5. This would make the vertex's ordered pairs (-2,-9/2). Step 3: Identify y-intercept and plot. The y-intercept is the "c"; therefore, the y-intercept is (0,-5/2). Step 4: Sketch. Place the y-intercept point on the other side of the axis of symmetry. This must have the same amount of space between each point for it to be a parabola and for it to be symmetrical. Then connect the points to have a "U" shape. Therefore, the lowest point or the minimum of the parabola is -9/2. |
Vertex Form-
Vertex form is y=a(x-h)^2+k, where the parent function is y=ax^2, The vertex is (h,k), where h will be the x and the k will be the y. The axis of symmetry is x=h. Your axis of symmetry allows you to plot symmetrical points. If a point is 2 places away from your axis of symmetry, the symmetrical point will be two places away on the other side of your axis of symmetry. You will know if your graph opens up or down by the number, a. If a is greater than 0, your graph will open up. If a is less than 0, it will open down.
How to graph in vertex form:
Steps:
- Find and plot the vertex (h,k)
- Figure out if your graph will open up or down.
- Plug in a x value to get a y value (Make sure it is a different x value from your vertex). Plot the point.
- Plot the symmetrical point.
- Sketch your graph.
Example:
Graph y=1(x+4)^2-5
Vertex=(4,-5)
a= 1, which is greater than 0, meaning our graph will open up.
x value= 1. Plug in x value
y=1(1+4)^2-5
y=1(5)^2-5
y=1(25)-5
y=25-5
y=20
So, plot (1, 20)
x value= 1. Plug in x value
y=1(1+4)^2-5
y=1(5)^2-5
y=1(25)-5
y=25-5
y=20
So, plot (1, 20)
Now we plot the symmetrical point. Since the point is three units away from the line of symmetry on the left, the symmetrical point will three points away on the right side, or six units away from the point, (1, 20).
So now, we sketch our graph. It should look like this:
And this is your answer.
Now, lets try to graph this equation: y= -1/4(x)^2-8
First, we need to find our vertex. Remember that our vertex is (h,k). Our h is 0 and our k is -8. So our vertex is (0,-8). So, lets plot that.
Now, lets try to graph this equation: y= -1/4(x)^2-8
First, we need to find our vertex. Remember that our vertex is (h,k). Our h is 0 and our k is -8. So our vertex is (0,-8). So, lets plot that.
And we also know that our graph will go down because our a is negative.
Now, lets plug in a x value. Lets put in 2.
y=-1/4(2)^2-8
y=-1/4(4)-8
y=-1-8
y=-9
So now we plot (2,-9)
Now, lets plug in a x value. Lets put in 2.
y=-1/4(2)^2-8
y=-1/4(4)-8
y=-1-8
y=-9
So now we plot (2,-9)
And now, we plot the symmetrical point. Since the point, (2, -9) is 2 units away from the axis of symmetry on the right side, the symmetrical point will be 2 units away on the left side, or 4 units away from (2, -9).
And then we sketch the graph.
And now you are done.
Now, a word problem.
A man is jumps across a hole. When he jumps, he starts from the edge and lands on the edge of the other side of the hole. The arc of his jump is represented by y=-1/8(x+4)^2+2, where it is all measured in feet. At what point does he jump the highest?
The highest point of the jump will be the vertex. Remember. To get your vertex, you will need h and k. h equals 4 and k equals 2. So, your vertex is (4, 2). Meaning, he jumped 2 feet in the air and that was the highest he jumped. And then you are done.
And that is Vertex Form.
Now, a word problem.
A man is jumps across a hole. When he jumps, he starts from the edge and lands on the edge of the other side of the hole. The arc of his jump is represented by y=-1/8(x+4)^2+2, where it is all measured in feet. At what point does he jump the highest?
The highest point of the jump will be the vertex. Remember. To get your vertex, you will need h and k. h equals 4 and k equals 2. So, your vertex is (4, 2). Meaning, he jumped 2 feet in the air and that was the highest he jumped. And then you are done.
And that is Vertex Form.
HOW TO USE YOUR CALCULATOR TO CHECK YOUR WORK-
Graph the function by first pressing the "y=" button above second then plug in the function and press graph on the calculator. Press second then trace which takes you to the calculate screen. Move your cursor down to maximum. Move you cursor left by using the direction pad and move it to the left of the maximum two clicks and hit enter, then move your cursor to the right of the maximum two click and hit enter. Put the cursor approximately on the maximum point and press enter. This gives you the best gues for the maximum "y" point. Then you have the solutions for the x value and y value.Practice Problems Vertex: Equation y= a(x – h)^2 +k
Graph the function by first pressing the "y=" button above second then plug in the function and press graph on the calculator. Press second then trace which takes you to the calculate screen. Move your cursor down to maximum. Move you cursor left by using the direction pad and move it to the left of the maximum two clicks and hit enter, then move your cursor to the right of the maximum two click and hit enter. Put the cursor approximately on the maximum point and press enter. This gives you the best gues for the maximum "y" point. Then you have the solutions for the x value and y value.Practice Problems Vertex: Equation y= a(x – h)^2 +k