Section 2.5 (cont): Stretches and Reflections
Last time, we saw how we could translate a graph by adding or subtracting to x or y:
UP by k
RIGHT by h
y = f (x) + k
y = f (x − h)
DOWN by k
LEFT by h
y = f (x) − k
y = f (x + h)
Now, to stretch or compress a graph, you MULTIPLY instead. Here are the rules for doing so:
STRETCH VERTICALLY by c
STRETCH HORIZONTALLY by c
y = cf (x)
y = f (x/c)
COMPRESS VERTICALLY by c
COMPRESS HORIZONTALLY by c
y = f (x)/c
y = f (cx)
Reflections: If c < 0 when you stretch, there is a reflection! For instance:
• y = f (−x/2): reflects horizontally (over y-axis) and stretches by a factor of 2.
• y = −f (x)/3: reflects vertically (over x-axis) and compresses vertically by a factor of 3.
Rules of thumb for all of these transforms
• Vertical transforms occur outside the function (i.e. they act on y), but horizontal transforms occur inside
(on x).
• Horizontal changes work in the “reverse” way from vertical.
• Perform horizontal changes first (i.e. handle inside before outside)!
Ex 1: Sketch a graph of each of the following:
(a) y = 4 − (x + 1)2 (Think of this as −(x + 1)2 + 4.)
(b) y = 2|x − 5| + 1
p
(c) y = − −x/2
Ex 2: For each part, a point P on the graph of
y = f (x) is given. For each transformation, determine
where the point moves.
(a) y = f (2x) − 1, P (2, 5)
(b) y = 1/5f (x − 1) + 3, P (0, −5)
Ex 3: In the graph, a parabola y = f (x) is drawn,
as well as three transforms of it. Find transformed
equations for each of the three shapes (a), (b), and (c).
(You may assume no stretches or compressions.)
Domain and Range Changes
Last time, we saw how range changes when you graph y = |f (x)| instead of f (x). Now, when you transform
a graph, you do the same changes to the domain and range! Remember:
Horizontal changes ⇒ Domain affected. Vertical changes ⇒ Range affected.
There is one other subtlety to note. When you do a reflection, you change the signs of the ends, so you also
flip their order!
• For instance, suppose y = f (x) has domain [−2, 3], and we study y = f (−x/2). We stretch horizontally by
2 to get [−4, 6]. Next, we reflect horizontally, so we get [−6, 4] (NOT [4, −6], which doesn’t make sense).
Ex 4: In each part, a domain D and range R is given for y = f (x). Find the new domain and range for
each transformation.
(a) y = 2f (x + 1), D = [0, 1], R = [3, 4]
(b) y = f (3x) + 1, D = [−6, 9], R = [−2, 4]
(c) y = f (−x/2)/2 − 3, D = [−2, 3], R = [0, 4]
(d) y = −3f (x + 4) + 1, D = [0, 4], R = [0, 4]
NOTE: A different way to handle these is to take the given domain and range and try drawing your own
function! Do the transforms on that, and then you can visually see the new domain and range.
Section 2.6A: Structure of Quadratic Functions
A quadratic function has the general form y = ax2 + bx + c where a 6= 0. Its shape is called a parabola.
The parabola points up when a > 0 and down when a < 0.
To solve a quadratic equation, you usually FIRST get everything to one side, leaving 0 on the other. Next,
you can either factor, or you can use the Quadratic Formula:
√
−b ± b2 − 4ac
2
ax + bx + c = 0 ⇒ x =
2a
Ex 5: Find the solutions for x in the following.
(a) 8x2 + 2 = x2 − 9x
(b) (x − 1)(x − 3) = 2
(c) −x2 + x − 3 = 0
NOTE: When solving, if a is negative, I usually multiply −1 to both sides to get a positive a instead.
Ex 6: Find the first-quadrant point where the parabola y = x2 + x intersects the line y = 2x + 2.
The Discriminant
The expression b2 − 4ac in the Quadratic Formula is the discriminant.
Sign of b2 − 4ac
Positive
Zero
Negative
# of Roots
2
1
0
How it looks
Graph crosses x-axis twice
Graph tangent to x-axis (glances it once)
Graph lies on one side of x-axis
Ex 7: How many roots do these quadratic functions have?
(a) x2 + 10x + 25
(b) 3x2 + 8x + 9
Ex 8: For which values of k does the formula kx2 − 6x + 3 have exactly one root?
Domain with Quadratics
Ex 9: Find the domain of each of the
√ following.
x+4
(b) g(x) = x2 − 6x + 8
(a) f (x) = x2 −2x+1
NOTE: For (b), you need to find out where x2 − 6x + 8 is positive or zero. First find the roots, then use the
roots to make a quick sketch of the graph... that’ll tell you which parts are positive.
Standard Form of a Quadratic
The point (h, k) where the parabola “turns around” is called the vertex of the parabola. When a > 0, the
vertex is the bottom of the graph. When a < 0, the vertex is at the top.
The standard form for a quadratic with leading coefficient a and vertex (h, k) is
y = a(x − h)2 + k
(To get this, start with y = ax2 and shift right by h and up by k!) Just like point-slope form for lines, the
standard form can often be obtained from word problems... you find the vertex, then solve for a.
Ex 10: (a) Find the vertex of y = −2(x + 5)2 − 3.
(b) Find the parabola with vertex (1, 3) which goes through (0, 5).
(c) Find the parabola with vertex (0, −9) with x-intercepts −2 and 2.
Converting to Standard Form: To convert y = ax2 + bx + c to y = a(x − h)2 + k, you normally complete
the square. However, we can do the following shortcut, known as the Vertex Formula:
1. The vertex x-coordinate is h =
−b
2a
.
2. Plug x = h back into the parabola to get y = k.
The a value is the same in both forms.
Ex 11: Find the vertex of each parabola.
(a) f (x) = 5 + 3x − x2
(b) g(x) = 2x2 + 9
(c) h(x) = 16x2 − 8x + 5