Respuesta :

Answer:

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

(b) [tex]y= -\dfrac12x^2+4x-6[/tex]

Step-by-step explanation:

(a)  From inspection of the graph, the x-intercepts (where [tex]y = 0[/tex]) are at

[tex]x = -2[/tex] and [tex]x =4[/tex].

Therefore [tex]y = a(x + 2)(x - 4)[/tex]   where [tex]a[/tex] is some constant

The y-intercept is at (0, -8)

Expanding the equation:

[tex]y = a(x + 2)(x - 4)[/tex]

[tex]\implies y = a(x^2-4x+2x-8)[/tex]

[tex]\implies y= ax^2-2ax-8a[/tex]

Therefore, -8a is the y-intercept

Comparing y-intercepts:  -8 = -8a  ⇒  a = 1

Substituting a = 1, the final equation in standard form is:

[tex]y=x^2-2x-8[/tex]

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(b)  From inspection of the graph, the x-intercepts (where [tex]y = 0[/tex]) are at

[tex]x = 2[/tex] and [tex]x =6[/tex].

Therefore [tex]y = a(x -2)(x - 6)[/tex]   where [tex]a[/tex] is some constant

The y-intercept is at (0, -6)

Expanding the equation:

[tex]y = a(x -2)(x - 6)[/tex]

[tex]\implies y = a(x^2-6x-2x+12)[/tex]

[tex]\implies y= ax^2-8ax+12a[/tex]

Therefore, 12a is the y-intercept

Comparing y-intercepts:

[tex]-6 = 12a \implies a = -\dfrac12[/tex]

Substituting found value of a into the equation:

[tex]\implies y= (-\dfrac12)x^2-8(-\dfrac12)x+12(-\dfrac12)[/tex]

[tex]\implies y= -\dfrac12x^2+4x-6[/tex]

y = a(x-h)²+k where (h,k) denotes the vertex

Q.1)

vertex: (1, -9) where h=1, k = -9

coordinates: (0, -8) where x = 0, y = -8

find for a:

y = a(x-h)²+k

-8 = a(0-1)²-9

-8 = a -9

a = 1

============

equation: y = (x-1)²-9

Q.2)

vertex: (4, 2) where h=4, k = 2

coordinates: (2, 0) where x = 2, y = 0

find for a:

y = a(x-h)²+k

0 = a(2-4)²+2

-2 = 4(a)

a = -1/2

============

equation: y = -1/2(x-4)²+2

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