Determine an equation of a Quadratic function
1 answer:
Answer:
y = -4x² + 32x - 48
Step-by-step explanation:
The standard form of a quadratic equation is
y = ax² + bx + c
We must find the equation that passes through the points:
(2, 0), (6,0), and (3, 12)
We can substitute these values and get three equations in three unknowns.
0 = a(2²) + b(2) + c
0 = a(6²) + b(6) + c
12 = a(3²) + b(3) + c
We can simplify these to get the system of equations:
(1) 0 = 4a + 2b + c
(2) 0 = 36a + 6b + c
(3) 12 = 9a + 3b + c
Eliminate c from equations (1) and (2). Subtract (1) from (2).
(4) 0 = 32a + 4b
Eliminate c from equations (2) and (3). Subtract (3) from (2).
(5) -12 = 27a - 3b
Simplify equations (4) and (5).
(6) 0 = 8a + b
(7) -4 = 9a - b
Eliminate b by adding equations (6) and (7).
(8) a = -4
Substitute (4) into (6).
0 = -32 + b
(9) b = 32
Substitute a and b into (1)
0 = 4(-4) + 2(32) + c
0 = -16 + 64 + c
0 = 48 + c
c = -48
The coefficients are
a= -4, b = 32, c = -48
The quadratic equation is
y = -4x² + 32x - 48
The diagram below shows the graph of your quadratic equation and the three points through which it passes.
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ANSWER:
0.25
i attached my work, hope this helps
0.25
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Answer:
The graphs of the two function will not intersect.
Step-by-step explanation:
We are given a quadratic function f(x).
Also g(x) is given by a set of values as:
x g(x)
1 -1
2 0
3 1
As g(x) is a linear function hence we find out the equation of g(x) by the slope intercept form of a line: y=mx+c
let g(x)=y
when x=1 , g(x)=-1
-1=m+c----(1)
when x=2 , g(x)=0
0=2m+c------(2)
Hence, on solving (1) and (2) by method of elimination we get:
m=1 and c=-2
Hence, the equation of g(x) is:
g(x)=x-2
So clearly from the graph we could see that the graph of the two functions will never intersect.
