1. Find the factored form of f(x) given it has x-intercepts (-3,0) and (4,0) and f(1)= -6. 2. Find the equation for f(x) given that it has roots of 2 and -3 and f(0)=12. Please someone help

1) f(x) = 0.5*(x + 3)*(x - 4)

2) f(x) = -2*x^2 - 2*x + 12

Step-by-step explanation:

Wen we have a quadratic function like:

f(x) = a*x^2 + b*x + c*x

And it has roots at x1 and x2, we can write the function in the factored form as:

f(x) = a*(x - x1)*(x - x2)

then:

1) we have x-intercepts at: (-3, 0) and (4, 0) (then the roots of this function are x = -3 and x =4) and we know that f(1) = -6

we have:

x1 = -3

x2 = 4

then we can write f(x) as:

f(x) = a*(x - (-3))*(x - 4)

f(x) = a*(x + 3)*(x - 4)

Where a is a real number.

and now we can use the fact that f(1) = -6

then:

f(1) = a*(1 + 3)*(1 - 4) = -6

       a*4*(-3) = -6

       a*-12 =  -6

       a = -6/-12 = 1/2 = 0.5

Then the function is:

f(x) = 0.5*(x + 3)*(x - 4)

2) Now we have roots x = 2 and x = .3

Then:

x1 = 2

x2 = -3

Then this function is something like:

f(x) = a*(x - 2)*(x - (-3))

f(x) = a*(x - 2)*(x + 3)

Now we know that f(0) = 12.

then:

f(0) = a*(0 - 2)*(0 + 3) = 12

         a*(-2)*3 = 12

          a*(-6) = 12

          a = 12/-6 = -2

f(x) = -2*(x - 2)*(x + 3)

And we do not want this one written in factored form, so we can just distribute the multiplications to et:

f(x) =  -2*(x - 2)*(x + 3) = (-2*x + 4)*(x + 3) = -2*x^2 + 4*x - 6*x + 12

f(x) = -2*x^2 - 2*x + 12

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐