Answer:
(a) [tex]y=(2x+6)(x-1)[/tex]
(b) [tex]y=2x^2+4x-6[/tex]
Step-by-step explanation:
Part (a)
The zeros are where the curve intersects the x-axis (where [tex]y = 0[/tex])
From the points given, the points of intersections are (-3, 0) and (1, 0)
[tex]x=-3\implies x+3=0[/tex]
[tex]x=1 \implies x-1=0[/tex]
Therefore, we can write the equation as:
[tex]y=a(x+3)(x-1)[/tex]
(where [tex]a[/tex] is some constant to be determined)
The y-intercept is when [tex]x = 0[/tex].
From the points given, the point of intersection with the y-axis is (0, -6)
Therefore, substitute [tex]x = 0[/tex] into the equation, set it to -6 and solve for [tex]a[/tex]:
[tex]\implies a(0+3)(0-1)=-6[/tex]
[tex]\implies a(3)(-1)=-6[/tex]
[tex]\implies -3a=-6[/tex]
[tex]\implies a=\dfrac{-6}{-3}=2[/tex]
Therefore, the final equation of the parabola is:
[tex]y=2(x+3)(x-1)[/tex]
in fully factored form: [tex]y=(2x+6)(x-1)[/tex] or [tex]y=(x+3)(2x-2)[/tex]
Part (b)
Standard form of a quadratic equation: [tex]y=ax^2+bx+c[/tex]
To express the equation in standard form, simply expand:
[tex]\implies y=(2x+6)(x-1)[/tex]
[tex]\implies y=2x^2-2x+6x-6[/tex]
[tex]\implies y=2x^2+4x-6[/tex]
