Quadratic functions

Question

The following functions are quadratic, either in standard form or general form.

F(x)=x^2-4x+3

G(x)=-(x-2)^2+4

H(x)=4x^2+1

 

For each function find:

A.) the coordinates of the vertex, V=(h,k)

B.) determine wheter the vertex is a minimum, or a maximun point

C.) the x- and y- intercepts

Answer 1

To find the vertex of each function, we can put them in vertex form:

 

y = a(x - h)² + k

 

where:

a is the coefficient of the x² term

V = (h, k)

 

 

1)

 

F(x) = (x² - 4x + 4) - 1

F(x) = (x - 2)(x - 2) - 1

F(x) = (x - 2)² - 1

 

 

From this form, we know the vertex is (2, -1).   Since the leading term is positive, this is a minimum point.

 

 

2)

 

G(x) is already in vertex form.

 

H(x) = -(x - 2)² + 4

 

Vertex is (2, 4).  Since the leading term is negative, this will be a maximum point.

 

 

3)

 

H(x) = 4(x + 0)² + 1

 

The vertex is (0, 1).  And since this parabola is positive, (0, 1) is a minimum.

 

 

 

To find the x-intercept for each function:

 

Set F(x), G(x), and H(x) equal to zero, and solve for x.

 

 

To find the y-intercept for each function:

 

Set x=0 and solve for F(x), G(x), and H(x).

 

 

 

Fun Fact:

 

Another way to find the vertex of a parabola is by understanding the transformations of  x².

 

 

(x - R)²  indicates a horizontal shift R units to the right from the origin.  The vertex starts at the origin.

 

(x + R)²  indicates a horizontal shift R units to the left from the origin.  The vertex starts at the origin.

 

-C  indicates a vertical shift C units down from the origin.  The vertex starts at the origin.

 

+C  indicates a vertical shift C units up from the origin.  The vertex starts at the origin.

Answer 2

First, I put them all in standard form: ax² + bx + c

 

F(x) = x² - 4x + 3

G(x) = -x² + 4x

H(x) = 4x² + 1

 

The vertex of a parabola is at h = -b / 2a. So these will be

 

F --> h = 2

G --> h = 2

H --> h = 0

 

Plug these back in to the function to get k = f(h):

 

F --> k = -1

G --> k = 4

H --> k = 1

 

You can tell if the vertex is a maximum or minimum by looking at a: if a > 0, the curve is sloping up so, it's a minimum; if a < 0, the curve is sloping down, so it's a maximum.

 

F --> (2,-1) [min]

G --> (2,4) [max]

H --> (0,1) [min]

 

To find the y-intercepts, look where x = 0:

 

F(x = 0) = 3

G(x = 0) = 0

H(x = 0) = 1

 

To find the x-intercepts, look where y = 0 (known as the roots). You could use the usual quadratic formula {x = (-b ± sqrt[b² - 4ac]) / 2a}, but these functions are simple enough that you don't have to.

 

F can be re-written as F(x) = (x - 1)(x - 3), so the function equals 0 when x = 1 or x = 3

 

G can be re-written as G(x) = x(4 - x), so the function equals 0 when x = 0 or x = 4

 

H doesn't have to be re-written: H(x) = 0 = 4x² + 1 --> x = ±sqrt[-1/4]. Since this is the square root of a negative number, there are no real roots.