Tutorial:
Graphing Quadratics

Question

Tutorial:
Graphing Quadratics

joel_the_boss · Answer

$$\Huge\text{Graphing Quadratics}\ $$

joel_the_boss · Answer

Before you learn how to graph these bad boys, your going to need to learn some basic things. :) 

 $$\Large\text{Parts of a Parabola}\ $$Parabolas consist of a vertex, axis of symmetry, domain, range, and sometimes x- and y-intercepts. Investigate each of these parts using the quadratic equation $$ y = -2(x - 5)^2 + 8 $$$$\Large\text{Vertex}\ $$ Finding the Vertex of a Parabola
•Step One: x-coordinate: Set the expression inside the parentheses equal to 0 and solve for the value of x.
•Step Two: y-coordinate: Substitute the value of x from step one into the original equation and solve for y.

joel_the_boss · Answer

Given the equation, $$f(x) = -2(x - 5)^2 + 8  $$,to algebraically find the x-coordinate of the vertex, set the expression x - 5 equal to 0 and solve for x.

joel_the_boss · Answer

|dw:1415449420470:dw|

joel_the_boss · Answer

To find the y-coordinate of the vertex, substitute the x-coordinate of the vertex into the original equation and solve for y.
$$y=-2(x-5)^2+8 $$ $$y=-2(5-5)^2+8 $$$$y=-2(0)^2+8 $$$$y=0+8 $$$$y=8$$

joel_the_boss · Answer

Therefore, the vertex of the parabola is (5, 8), which is identical to (h, k). Since a is negative, the parabola will open down and the vertex will represent the maximum (or highest) point on the graph, as you can see below.

joel_the_boss · Answer

$$\Large\text{Axis of Symmetry}\ $$ Using the graph of the equation, $$f(x) = -2(x - 5)^2 + 8 $$ again, create a line through the parabola so that if the parabola were to be folded down that line, the two sides of the parabola would completely match.

joel_the_boss · Answer

$$\Large\text{Domain and Range}\ $$The domain is the group of numbers that can be substituted for the variable x and produce a unique value for y. Take another look at the graph of f(x) = -2(x - 5)2 + 8.

No matter what number is substituted for x in the quadratic equation, $$ f(x) = -2(x - 5)^2 + 8,$$there will be a resulting y value that is unique. Therefore, the domain of this equation is "all real numbers."

The range is the group of numbers that can be produced from all of the domain values. In other words, the range is the y values that have a corresponding x value on the parabola.

The vertex of the parabola is (5, 8). The parabola opens down so the vertex represents the maximum point on the parabola. All of the y values must be below that y-coordinate of 8. Therefore, the range of this equation is y ≤ 8.

joel_the_boss · Answer

$$\Large\text{Intercepts}\ $$So far, you have discovered how to find the vertex, axis of symmetry, domain, and range of parabolas both algebraically and graphically. The x- and y-intercepts of a quadratic equation may also be determined algebraically and graphically. However, the algebraic process of finding x-intercepts will be explored in a later lesson. Fortunately, by looking at the graph, it is sometimes easy to see where the parabola crosses the x-axis. *Look at attachment 1* 

Since this parabola crosses the x-axis in two places, there are two x-intercepts: (3, 0) and (7, 0). By scrolling the parabola down, you can also find the point where the graph crosses the y-axis. *Attachment 2*

The parabola crosses the y-axis at one point: (0, -42). Therefore, this single point is the only y-intercept for the graph. You can find this value algebraically by substituting 0 for x in the equation and solving for y.

joel_the_boss · Answer

$$f(x) = -2(x - 5)^2 + 8$$$$f(x) = -2(0 - 5)^2 + 8 $$$$f(x) = -2(-5)^2 + 8$$$$f(x) = -2(25) + 8  $$$$f(x) = -50 + 8  $$ $$f(x) = -42 $$

joel_the_boss · Answer

$$\Large\text{Standard Form of a Quadratic Equation} \ $$While identifying the vertex, axis of symmetry, domain, range, and intercepts is most easily done using the general form of a quadratic equation $$ [f(x) = a(x - h)2 + k]$$Most of the time quadratic equations are written in standard form. It is very important to know how to find the parts of a parabola regardless of the form, general or standard.  

The standard form of a quadratic equation, which is$$ f(x) = ax^2 + bx + c $$, can easily be found by expanding the general form of a quadratic equation.

joel_the_boss · Answer

As you may have noticed, the numerical value in the equation of the axis of symmetry was always equal to the x-coordinate of the vertex. So by finding the equation of the axis of symmetry, you're halfway to finding the vertex!

joel_the_boss · Answer

$$y=(x−5)^2−4  $$Identify the vertex, equation for the axis of symmetry, domain, range and x- and y-intercepts of the parabola.

From the general form of the parabola,$$y=a(x−h)^2+k, $$the vertex is given by (h,k). So in this case, the vertex is (5,−4).

The axis of symmetry is found by setting x−5=0, giving you x=5.

Because any real number can be substituted for x in the function, the domain is all real numbers.

Since a is positive, the parabola opens up. The minimum value of the parabola occurs at y=−4; therefore the range is all real numbers greater than or equal to −4.

The x-intercepts occur where y=0 and solving for x. Because the solutions to $$0=(x−5)^2−4,$$ are x=3 and x=7, the x-intercepts are (3,0) and (7,0). The y-intercept occurs where x=0 and solving for y. Because $$(0−5)^2−4=21 $$ , the y-intercept is (0,21).

thesmartone · Answer

Good Job! @Joel_the_boss