PSDA - Graphs of Linear, Quadratic and Exponential Functions Lesson

Graphs of Linear, Quadratic and Exponential Functions

Linear vs. Quadratic vs. Exponential Functions

Graphing Quadratic and Exponential Functions means that curvature is added to the graphs. All the points of a linear function lay on the straight line that could be drawn between any of the line's two points. A linear function is considered a function of degree 1 since the x value occurs only once. Remember that the linear function was of the form y = mx + b, where m is the slope and b is the y-intercept.

Linear Function

An equation of degree 1, which has at most 1 solution. The slope intercept form of the equation is y = mx + b, where m is the slope and b is the y-intercept. Known as a straight line.

Quadratic Function

An equation of degree 2, which has at most two solutions. The general form of a quadratic equation is where a and b are coefficients (numbers multiplied by the x term) and c is the constant (a number). The y-intercept is c. Graphs of these are called parabolas.

Exponential Function

A function of the form where both a and b are numbers not equal to 0 and b is not equal to 1. This function never crosses the x-axis to have solutions, but gets closer and closer to the x-axis.

A change to the form to where c is a constant, moves the graph up and down the y-axis, just like the b of the linear equation and a + c in this equation is the y-intercept.

y = x
graph
Linear f(x) = x

y = x^2
graph
Quadratic f(x) = x^2

y = 2^x
graph
Exponential f(x) = ab^x

A function is an equation that has one value of y for each x-value. Function notation substitutes y = f(x). The equations under the graphs are written in function notation.

y=-x of f(x) = - x
y= -x^2 or f(x) = -x^2
y = 2^x or f(x) = -2^x

Using this notation, f(2) = x + 1 means to substitute x = 2 and solve the equation. Solving the equation f(2) = 2 + 1 = 3, giving the point on the graph (2, 3).

The vertex of the parabolic equation is the absolute maximum or absolute minimum point on the function's graph. In the graph the vertex is seen to be (0, 0), the absolute minimum. The graph of would also have the vertex at (0, 0) but the vertex would be an absolute maximum as the graph is reflected over the x-axis with the negative. Look at the graphs below. They are reflections of the graphs above across the x-axis due to the negative being in front of the x squared term. equation image indicator

Note that the negative sign is not squared with the x-value. For the negative sign to be squared with the x-value, the equation would need to be written as f(x) = (-x)^2. This is an important concept of exponent powers to remember. Here is the difference:

$LaTeX: 3^2=9\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(-3\right)^2=9\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-3^2=-9$ $3^2=9\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(-3\right)^2=9\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-3^2=-9$ the 3 is squared, not the negative

$LaTeX: 3\cdot3=9\:\:\:\:\:\:\:\:\:\:\left(-3\right)\left(-3\right)=9\:\:\:\:\:\:\:\:\:\:-\left(3\right)\left(3\right)=-9$ $3\cdot3=9\:\:\:\:\:\:\:\:\:\:\left(-3\right)\left(-3\right)=9\:\:\:\:\:\:\:\:\:\:-\left(3\right)\left(3\right)=-9$ negative times 3 times 3

Linear vs. Quadratic vs. Exponential Functions

Type of Function	Linear	Quadratic	Exponential
Equation 1	y = mx + b Slope Intercept m = slope b = constant	$y=ax^2+bx+c$ General Form a & b = coefficients c = constant axis of symmetry is $LaTeX: x=-\frac{b}{2a}$ $x=-\frac{b}{2a}$ vertex = $LaTeX: \left(-\frac{b}{2a},\:f\left(-\frac{b}{2a}\right)\right)$ $\left(-\frac{b}{2a},\:f\left(-\frac{b}{2a}\right)\right)$	$y=ab^x+c$ a = coefficient b = base c = horizontal asymptote (y = c) the asymptote is not crossed
Equation 2	$LaTeX: y_1-y_2=m\left(x_2-x_1\right)$ $y_1-y_2=m\left(x_2-x_1\right)$ find the equation given two points	$LaTeX: y=a\left(x-h\right)^2+k$ $y=a\left(x-h\right)^2+k$ Vertex Form Vertex (h, k)	Analysis of b b > 1, graph is increasing; 0 < b < 1, graph is decreasing
y-intercept	b in slope-intercept form	c in general form or plug in x = 0 in either form and solve for y	a + c, plug in x = 0; a is a number and b⁰= 1
Number of solutions (where the x-axis is crossed).	1 at most Horizontal lines do not cross the x-axis	2 at most	None unless shifted to cross the x-axis
Find solutions (x-intercepts)	Using slope-intercept form with known slope, plug in y = 0 and solve for x	Using general form knowing a, b, and c, plug in y = 0 and solve for x	Sometimes, must be shifted vertically to cross the x-axis.
Domain	All real numbers	All real numbers	All real numbers
Range	All real numbers	y ≥ k if upward parabola or y ≤ k if downward parabola (negative sign in front of a)	y > c if concave up or y < c if concave down (negative sign in front of a)

Example 1

Graph the equation LaTeX: y=2x^2-8 $y=2x^2-8$

Further explanation of this problem will be found by watching the video. Let's work this together and discover how these answers were obtained.

The graph will be an upward parabola because the 2 in front of the x squared does not have a negative sign.

Method 1: Create a table of values -3 < x < 3 find the y-values and plot the points by hand, connecting the curve.

x	y
-3	y = 2(-3)² - 8 = 10
-2	y = 2(-2)²- 8 = 0
-1	y = 2(-1)² - 8 = -6
0	y = 2(0)² - 8 = -8
1	y = 2(1)² - 8 = -6
2	y = 2(2)² - 8 = 0
3	y = 2(3)²- 8 = 10

Method 2: Recognizing the y-intercept

Using this method, the vertex is solved for.

Input two other easy x-values to find two other points on the same side of the vertex, reflect across the y-axis, and sketch the curve.

The equation is in general form with b = 0.

Plugging in x= 0 in the equation yields y = -8. So the vertex of the parabola is (0, -8).

Method 3: Find the axis of symmetry and then solve for the vertex.

Axis of Symmetry: Solve for -b/(2a) = -0/(2*2) = 0

Solve for f(0) = 2(0)^2 - 8 = 8

The vertex is (0, -8)

Input into the equation two other easy x-values to find two other points on the same side of the vertex, reflect across the y-axis, and sketch the curve.

Graph: Domain is all real numbers

Range is y ≥ -8

Upward parabola

y-intercept = -8

x-intercepts = -2 & 2

symmetric y-axis is x = 0

Image parabola
Range is y ≥ -8

Upward parabola

y-intercept = -8

x-intercepts = -2 & 2

symmetric y-axis is x = 0

Example 2

Exponential graphs rise, are increasing, concave up, from left to right if their base is greater than 1, b>1. If their base is between 0 and 1 (a positive fraction < 1), then the exponential flips across the y-axis and is decreasing, concave up.

Lastly, a number constant c on the end of the equation y = ab^x + c, shifts the graph up and down. This changes the horizontal asymptote from y = 0 (understood in the original graphs as no value of c was shown) to the value of c. See the example below.

Describe the exponential graph $LaTeX: y=\left(\frac{1}{2}\right)^x+1$ $y=\left(\frac{1}{2}\right)^x+1$

Graph: opens upward, concave up

y-intercept is (1/2)⁰ + 1 = (1)+1 = 2

x-intercept: none

base < 1: decreasing left to right, concave up (the base is the 1/2)

Domain: all real numbers

Range: y > 1 (shifted up 1 unit)

Image concave up
opens upward, concave up

y-intercept is (1/2)^0 + 1 = (1)+1 = 2

x-intercept: none

base < 1: decreasing left to right, concave up (the base is the 1/2)

Domain: all real numbers

Range: y > 1 (shifted up 1 unit)

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