We are given the points
<span>(0,-4),(1,0),(2,2)
The standard form a quadratic function (in terms of x) is
y = Ax2 + Bx + C
Subsitute the points
-4 = 0 + 0 + C
C = -4
0 = A + B - 4
2 = 4A + 2B - 4
Solve for A and B
A = -1
B = 5
The function is
y = -x2 + 5x - 4</span>
Answer:
x is less than -3
Step-by-step explanation:
3x^2+9x<0
x^2+3x<0
x(x+3)<0
x+3<0
x<-3
Answer:
x = -0.6
y = 2.2
z = 2
Step-by-step explanation:
2x + y - 2z = -3
x + 3y - z = 4
3x + 4y - z = 5
Rewrite the system in matrix form and solve it by Gaussian Elimination (Gauss-Jordan elimination)
2 1 -2 -3
1 3 -1 4
3 4 -1 5
R1 / 2 → R1 (divide the 1 row by 2)
1 0.5 -1 -1.5
1 3 -1 4
3 4 -1 5
R2 - 1 R1 → R2 (multiply 1 row by 1 and subtract it from 2 row); R3 - 3 R1 → R3 (multiply 1 row by 3 and subtract it from 3 row)
1 0.5 -1 -1.5
0 2.5 0 5.5
0 2.5 2 9.5
R2 / 2.5 → R2 (divide the 2 row by 2.5)
1 0.5 -1 -1.5
0 1 0 2.2
0 2.5 2 9.5
R1 - 0.5 R2 → R1 (multiply 2 row by 0.5 and subtract it from 1 row); R3 - 2.5 R2 → R3 (multiply 2 row by 2.5 and subtract it from 3 row)
1 0 -1 -2.6
0 1 0 2.2
0 0 2 4
R3 / 2 → R3 (divide the 3 row by 2)
1 0 -1 -2.6
0 1 0 2.2
0 0 1 2
R1 + 1 R3 → R1 (multiply 3 row by 1 and add it to 1 row)
1 0 0 -0.6
0 1 0 2.2
0 0 1 2
x = -0.6
y = 2.2
z = 2
It's just wherever the function touches the x axis, so 5 and -1
Answer:
is the required equation.
Therefore, the second option is true.
Step-by-step explanation:
We know that the slope-intercept form of the line equation of a linear function is given by

where m is the slope and b is the y-intercept
Taking two points (0, -2) and (1, 0) from the table to determine the slope using the formula




substituting the point (0, -2) and the slope m=2 in the slope-intercept form to determine the y-intercept i.e. 'b'.




Now, substituting the values of m=2 and b=-2 in the slope-intercept form to determine the equation of a linear function



Thus,
is the required equation.
Therefore, the second option is true.