We know from the first point that (x -2) is a factor of the polynomial, so we can write the other factor as (ax+b). Filling in the values from the other two given points, we have ...
f(x) = (x -2)(ax +b)
f(3) = (3 -2)(3a +b) = 2
f(4) = (4 -2)(4a +b) = 6
__
From the first of these, ...
3a + b = 2
Dividing the second by 2, we have ...
4a + b = 3
Subtracting the first of these equations from the second gives ...
(4a +b) -(3a +b) = (3) -(2)
a = 1
Using this in the first of the above equations, we have ...
3·1 + b = 2
b = -1
Then the factored form of the equation is ...
f(x) = (x -2)(x -1)
Expanding this to standard form, we have ...
f(x) = x² -3x +2
[tex]What is the equation, in standard form, of a parabola that contains the following points? (2,0), (3[/tex]
D
Step-by-step explanation:
the the 5 minus the 12 and the 0 minus the negative one all finds it way. back to the equation d
-20 = 4a - 2b + c
-4 = c
-20 = 16a + 4b + c
4a - 2b + c = 16a + 4b + c
4a - 2b = 16a + 4b
-6b = 12a
b = -2a
-20 = 4a - 2b + c
-20 = 4a + 4a - 4
-16 = 8a
-2 = a
b = 4
c = -4
y = -2x^2 + 4x - 4
y = -2 * (x^2 - 2x) - 4
y = -2 * (x^2 - 2x + 1 - 1) - 4
y = -2 * (x^2 - 2x + 1) + 2 - 4
y = -2 * (x + 1)^2 - 2
Y = -2.5x^2 + 4.5^x + -4.5
[tex]What is the equation, in standard form, of a parabola that contains the following points? ( -2, -23[/tex]
y = -2*3x - 1
Step-by-step explanation:
The first term is -2 and the common ratio is 3 because every term is being multiplied by 3. The equation is y = -2*3x - 1.
The equation of parabola is [tex]y=-\frac{5}{2}x^2+\frac{9}{2}x-4.5[/tex].
Step-by-step explanation:
Equation of a parabola is quadratic equation.
Let the equation of parabola be
[tex]y=Ax^2+Bx+C[/tex] .... (1)
The parabola contains points ( -2, -23.5), (0,-4.5), (4,-26.5).
Put (0,-4.5) in equation (1),
[tex]-4.5=A(0)^2+B(0)+C[/tex]
[tex]-4.5=C[/tex]
Put this value in equation (1).
[tex]y=Ax^2+Bx-4.5[/tex] ... (2)
Put ( -2, -23.5) and (4,-26.5) in equation (2).
[tex]-23.5=A(-2)^2+B(-2)-4.5[/tex]
[tex]-19=4A-2B[/tex] .... (3)
[tex]-26.5=A(4)^2+B(4)-4.5[/tex]
[tex]-22=16A+4B[/tex] .... (4)
On solving (3) and (4), we get
[tex]A=-\frac{5}{2}[/tex] and [tex]B=\frac{9}{2}[/tex]
Therefore the value of parabola is
[tex]y=-\frac{5}{2}x^2+\frac{9}{2}x-4.5[/tex]
[tex]What is the equation, in standard form, of a parabola that contains the following points? ( -2, -23[/tex]
I believe your answer is C) y = -2x^2 + 4x - 4
C) y=x^2-2x+6
Step-by-step explanation:
We are given three points
(1,5) (-1,9) and (4,14)
We can verify each options
option-A:
[tex]y=x^2+6x+2[/tex]
we will verify each points
At (1,5):
we can plug x=1 and check whether y=5
[tex]y=(1)^2+6(1)+2[/tex]
[tex]y=9[/tex]
It does not satisfy point
So, this is FALSE
option-B:
[tex]y=x^2+6x-2[/tex]
we will verify each points
At (1,5):
we can plug x=1 and check whether y=5
[tex]y=(1)^2+6(1)-2[/tex]
[tex]y=5[/tex]
It satisfies point
At (-1,9):
we can plug x=-1 and check whether y=9
[tex]y=(-1)^2+6(-1)-2[/tex]
[tex]y=-7[/tex]
It does not satisfy point
So, this is FALSE
option-C:
[tex]y=x^2-2x+6[/tex]
we will verify each points
At (1,5):
we can plug x=1 and check whether y=5
[tex]y=(1)^2-2(1)+6[/tex]
[tex]y=5[/tex]
It satisfies point
At (-1,9):
we can plug x=-1 and check whether y=9
[tex]y=(-1)^2-2(-1)+6[/tex]
[tex]y=9[/tex]
So, it satisfies point
At (4,14):
we can plug x=4 and check whether y=14
[tex]y=(4)^2-2(4)+6[/tex]
[tex]y=14[/tex]
So, it satisfies point
so, this is TRUE
C) y = -2x^2 +4x - 4
Step-by-step explanation:
The y-values are the same for points (-2, -20) and (4, -20), so the axis of symmetry is halfway between those points, at x = (-2+4)/2 = 1.
The y-intercept is (0, -4), so the only viable answer choices are B and C. The axis of symmetry is given by ...
x = -b/(2a)
For choice B, this is x = -4/(2(-1)) = 2 (doesn't work).
For choice C, this is x = -4/(2(-2)) = 1, which matches the above analysis.
The appropriate choice is ...
y = -2x^2 +4x - 4
_____
Alternate solution
If you like, you can derive the equation for the parabola. Since you know that the y-intercept is -4, you can write the equation as ...
y = ax² +bx -4
Filling in the data points that are not x=0, we have two equations in two unknowns:
-20 = a(-2)² +b(-2) -4 ⇒ 4a -2b = -16
-20 = a(4)² + b(4) -4 ⇒ 16a +4b = -16
Adding twice the first equation to the second gives ...
2(4a -2b) + (16a +4b) = 2(-16) +(-16)
24a = -48
a = -2 . . . . . . . . matches choice C
4(-2) -2b = -16 . . . . . substitute into an equation to find b
-2b = -8 . . . . . . . . . . add 8
b = 4 . . . . . . . . . . . . . divide by -2
The equation that fits the given data is ...
y = -2x² +4x -4
[tex]What is the equation, in standard form, of a parabola that contains the following points? (-2, -20)[/tex]
answer:
15
step-by-step explanation:
f(x) = x² -3x +2
Step-by-step explanation:
We know from the first point that (x -2) is a factor of the polynomial, so we can write the other factor as (ax+b). Filling in the values from the other two given points, we have ...
f(x) = (x -2)(ax +b)
f(3) = (3 -2)(3a +b) = 2
f(4) = (4 -2)(4a +b) = 6
__
From the first of these, ...
3a + b = 2
Dividing the second by 2, we have ...
4a + b = 3
Subtracting the first of these equations from the second gives ...
(4a +b) -(3a +b) = (3) -(2)
a = 1
Using this in the first of the above equations, we have ...
3·1 + b = 2
b = -1
Then the factored form of the equation is ...
f(x) = (x -2)(x -1)
Expanding this to standard form, we have ...
f(x) = x² -3x +2
[tex]What is the equation, in standard form, of a parabola that contains the following points? (2,0), (3[/tex]