d The graph of h is a horizontal translation 6 units to the right of the graph of f.
A
aDomain: All real numbers
Range: y≤ 8 Zeros: x=3 and x=7
B
b y=- 2x^2+20x-42
C
c See solution.
D
dGraph:
Practice makes perfect
a Let's analyze the given graph.
The domain of any quadratic function is all real numbers. Since the parabola opens downwards and its vertex is (5,8), the range of f is y≤ 8. From the given graph we can see that the parabola meets the x-axis at x=3 and x=7, so these two points are zeros of this function.
b First we will use the vertex form of a quadratic function.
y=a(x- h)^2+ k
Here ( h, k) are coordinates of the vertex. In our case, ( h, k)=( 5, 8).
y=a(x- h)^2+ k
⇓
y=a(x- 5)^2+ 8
Now, we will substitute (x,y)=(3,0) and solve the equation to find a.
c First, let's plot y=x^2 and y=f(x) in the same coordinate system.
The graph of f opens downward and is 2 times narrower than the graph of y=x^2. The vertex of the graph f, which is ( 5, 8), is 5 units right and 8 units up of the vertex of y=x^2, which is (0,0). So, the graph of f is a vertical stretch by a factor of 2, a reflection in the x-axis, and a translation 5 units right and 8 units up of the graph of y=x^2.
d Let's figure out what transformations were involved in the new function so we can graph it.
Transformations of y=f(x)
Vertical Translations
Translation up k units, k>0
y=f(x)+ k
Translation down k units, k>0
y=f(x)- k
Horizontal Translations
Translation right h units, h>0
y=f(x- h)
Translation left h units, h>0
y=f(x+ h)
Vertical Stretch or Compression
Vertical stretch, a>1
y= af(x)
Vertical compression, 0< a<1
y= af(x)
Horizontal Stretch or Compression
Horizontal stretch, 0< b<1
y=f( bx)
Horizontal compression, b>1
y=f( bx)
Reflections
In the x-axis
y=- f(x)
In the y-axis
y=f(- x)
We want to graph h(x)=f(x-6). Looking at the table, we can say that the graph of f(x) is translated 6 units to the right. With this information, let's graph the function!
