Question

A parabola has the vertex (4.5, –10).
a) Write an equation to describe all parabolas with this vertex.
b) A parabola with the given vertex passes through the point (–1, 0). Determine the equation for this parabola.
c) State the domain and range of the function

Answer 1

Parabola equation is y = ax^2+bx+c

Vertex form is y = a(x-h)^2+k

a) (h,k) = (4.5,-10)

y = a(x-4.5)^2-10

b) (h,k) = (-1,0)

y = a(x+1)^2+0

y = a(x+1)^2

y = a(x^2+1+2x)

y = ax^2+a+2ax

y = ax^2+2ax+a -------> (1)

Compare with ax^2+bx+c = 0

a = a , b = 2a, c = a

axis of symmetry x = -b/2a

x = -2a/2a

x = -1

y = a(-1)^2+2(-1)+c = a-2+c

Now substitute x ,y in (1).

a-2+c = a(-1)^2+2a(-1)+c

a-2+c = a-2a+c

Bring all terms to one side.

a-2+c-a+2a-c = 0

2a-2 = 0

Add 2 to each side.

2a = 2

a = 1

Now substitute a in (1)

y = 1*x^2+2*1*x+1

Equaton of parabola is y = x^2+2x+1.

Answer 2

Vertex form of parabola  y  = a (x - h ) ²+ k

Where (h ,k ) is the vertex.

a) Parabola vertex at (4.5 , -10) = (h , k )

Equation of parabola with vertex (4.5, -10) is y  = a (x - 4.5)²  - 10

When a  is positive number the parabola opens upward.

When a  is negitive number the parabola opens downward.

b) The parabola vertex (4.5,-10) and passes through (-1,0).

y  = a (x - 4.5)²  - 10

To find a  value

Substitute (x, y) = (-1,0) in y = a (x - 4.5)²  - 10.

0 = a (-1 - 4.5)²  - 10

0 = 30.25a -10

30.25a = 10

a = 10/30.25 = 0.33

The equation is y  = 0.33 (x - 4.5)²  - 10.

c) To find domain and range of the y = 0.33 (x - 4.5)²  - 10

The above function represents a parabola vertex form  y = a (x - h )² + k .

 a  = 0.33 , h  = 4.5 and k  = -10.

a  is positive number the parabola opens up and has minimum value.

When the parabola opens up it has a minimum point which is the vertex of parabola (4.5, -10)

In the minimum point y  = -10  so the graph of parabola cannot be lower than -10.

Thus the range of function y ≥ -10.

Domain of function is all real numbers.

Range of the function is .