What is the quadratic function to these three points (-1,-11), (0,-3), and (3,-27)

y = -4x² + 4x -3

Step-by-step explanation:

The standard form of quadratic equation is:

y = ax² + bx + c

We need to use the points given to find the value of a, b and c

We are given point (-1,-11) where x =-1 and y=-11 putting values in the above equation

y = ax² + bx + c

-11 = a(-1)² + b(-1) +c

-11 = a -b+c      eq(1)

Now putting the point(0, -3) where x =0 and y =-3

y = ax² + bx + c

-3 = a(0)² + b(0) + c

-3 = 0 + 0 + c

=> c = -3

Now Putting the point (3, -27) where x =3 and y = -27

y = ax² + bx + c

-27 = a(3)² +b(3) + c

-27 = 9a + 3b + c  eq(2)

Putting value of c= -3 in eq(2)

-27 = 9a + 3b -3  

-27 +3 = 9a +3b

-24 = 9a + 3b

=> 3(3a+b) = -24

3a + b = -24/3

3a + b = -8     eq(3)

Putting value of c= -3 in eq(1)

-11 = a -b+c

-11 = a -b -3

-11 + 3 = a - b

-8 = a - b    

=> a - b = -8   eq(4)

Now adding eq(3) and eq(4)

3a + b = -8  

a - b = -8

4a = -16

a = -16/4

a = -4

Putting value of a in equation 4

a - b = -8

-4 -b = -8

-b = -8+4

-b = -4

=> b = 4

The values of a , b and c are a= -4, b =4 and c= -3

Putting these values in standard quadratic equation

y = ax² + bx + c

y = -4x² + 4x -3

you have 6 cupcakes

the formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. if a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.