Write the quadratic function in three forms using the three points it goes through (-1,24),(5,0), and (-4,18)

The Standard form becomes - y = x² x +

Step-by-step explanation:

To find - Write the quadratic function in three forms using the three points it goes through (-1,24),(5,0), and (-4,18)

Proof -

There are there forms of quadratic equation :

1) Standard form: y = ax²+ bx + c

where the a,b, and c are constants

2) Factored form: y = (ax + c)(bx + d)

where a,b,c, and d are constants

3) Vertex form: y = a(x + b)² + c

where a, b, and c are constants

Now,

For Standard form -

Given that the quadratic equation passes through three points

(-1, 24), (5, 0), (-4, 18)

24 = a(-1)²+ b(-1) + c

⇒24 = a - b + c                       ............(1)

0 = a(5)²+ b(5) + c

⇒0 = 25a + 5b + c                       ............(2)

18 = a(-4)²+ b(-4) + c

⇒18 = 16a - 4b + c                       ............(3)

By solving equation (1), (2), (3), we get

a = , b = , c =

So,

The Standard form becomes - y = x² x +

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92