GUYS PLEASE I NEED HELP

Given:

The graphs of two parabolas.

To find:

The equation of the quadratic function with given graphs.

Solution:

(a)

The factor form of a parabola is

[tex]y=a(x-b)(x-c)[/tex]

Where, a is a constant, b and c are x-intercepts.

From the graph (a) it is clear that the graph intersect the x-axis at -1 and 3. So, b=-1 and c=3.

[tex]y=a(x-(-1))(x-3)[/tex]

[tex]y=a(x+1)(x-3)[/tex]            ...(i)

Put x=0 and y=3 because the y-intercept is (0,3).

[tex]3=a(0+1)(0-3)[/tex]

[tex]3=-3a[/tex]

[tex]\dfrac{3}{-3}=a[/tex]

[tex]-1=a[/tex]

Putting a=-1 in (i), we get

[tex]y=-1(x+1)(x-3)[/tex]

[tex]y=-(x+1)(x-3)[/tex]

Therefore, the equation of the parabola is [tex]y=-(x+1)(x-3)[/tex].

(b)

The vertex form of a parabola is

[tex]y=a(x-h)^2+k[/tex]

Where, a is a constant and (h,k) is vertex.

From the the graph (b), it is clear that the vertex of the of the parabola is at point (2,0) and y-axis is (0,8). So, h=2, k=0.

[tex]y=a(x-2)^2+0[/tex]

[tex]y=a(x-2)^2[/tex]            ...(ii)

Put x=0 and y=8 because the y-intercept is (0,8).

[tex]8=a(0-2)^2[/tex]

[tex]8=4a[/tex]

[tex]\dfrac{8}{4}=a[/tex]

[tex]2=a[/tex]

Putting a=2 in (ii), we get

[tex]y=2(x-2)^2[/tex]

Therefore, the equation of the parabola is [tex]y=2(x-2)^2[/tex].

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