Question

Match each graph with its corresponding equation

Answer 1

The graphs match the equations as follows:

Graph A: (−x+3)^2 -2

Graph B: (x−2)^2+3

Graph C : −(x+2)^2 +3

Graph D : 2(x−2)^2 +3

Equation (−x+3)^2 -2:

This equation is in the form of a quadratic function, y=a(x−h) ^2+k, where (h ,k) is the vertex of the parabola.

The vertex of the parabola is at (3,1), which matches the vertex of Graph A.

The coefficient a is negative, which tells us that the parabola opens downward.

The constant term k is −2, which tells us that the minimum value of the parabola is −2.

Equation (x−2)^2+3 :

This equation is in the form of a quadratic function, y=a(x−h) ^2 +k, where (h, k) is the vertex of the parabola.

The vertex of the parabola is at (2,3), which matches the vertex of Graph B. The coefficient a is positive, which tells us that the parabola opens upward.

The constant term k is 3, which tells us that the minimum value of the parabola is 3.

Equation −(x+2)^2 +3:

This equation is in the form of a quadratic function, y=a(x−h) ^2 +k, where (h, k) is the vertex of the parabola.

The vertex of the parabola is at (−2,3), which matches the vertex of Graph C.The coefficient a is negative, which tells us that the parabola opens downward.

The constant term k is 3, which tells us that the maximum value of the parabola is 3.

Equation 2(x−2)^2 +3 :

This equation is in the form of a quadratic function, y=a(x−h) ^2 +k, where (h ,k) is the vertex of the parabola.

The vertex of the parabola is at (2,3), which matches the vertex of Graph D. The coefficient a is positive and multiplied by 2, which tells us that the parabola opens upward and is stretched vertically by a factor of 2.

The constant term k is 3, which tells us that the minimum value of the parabola is 3.

Answer 2

The equation of the parabolas given will be found as follows:
a] general form of the parabolas is:
y=k(ax^2+bx+c)
taking to points form the first graph say (2,-2) (3,2), thus
y=k(x-2)(x-3)
y=k(x^2-5x+6)

taking another point (-1,5)
5=k((-1)^2-5(-1)+6)
5=k(1+5+6)
5=12k
k=5/12
thus the equation will be:
y=5/12(x^2-5x+6)

b] Using the vertex form of the quadratic equations:
y=a(x-h)^2+k
where (h,k) is the vertex
from the graph, the vertex is hence: (-2,1)
thus the equation will be:
y=a(x+2)^2+1
taking the point say (0,3) and solving for a
3=a(0+2)^2+1
3=4a+1
a=1/2
hence the equation will be:
y=1/2(x+2)^2+1