First question (sorry if in wrong section)

[Sirus]

hey guys this is my first time on these forums so sorry if this post is in the wrong forum section. But i was doing a question on tables equations and graphs and one question was involving the translation of a parabola. on the right is the marking schedule aka the answers while on the left is my working out. I don't think the marking schedule is correct. So is my answer right or is the marking schedule correct?

 

[Sirus]

BTW the answer on the schedule is a(iii)

 

[hoosie]

Transformation of parabola y = x^2 -3x -4

Parabola shown cuts y-axis at -4 and x-axis at x = -1 or x = 4.
Equation: y = a(x + 1)(x -4)
We don't know the value of 'a' - the dilation factor.
We will assume a = 1 and work out the minimum y- value of the parabola to
see if it coincides with the value shown on the graph (a little less than -6).
y = (x + 1)(x -4)
= x² - 3x - 4
Axis of symmetry is half-way between x = -1 and x = 4.
Therefore x = 3/2 is the equation of this line.


Work out y-value of TP by substituting x = 3/2 into equation:
y = (x + 1)(x - 4)
= (3/2 + 1)(3/2 - 4)
= 5/2 * -5/2
= -25/4
Graph turns just below -6 so this tells us a = 1.


Write rule in TP form:
y = (x - 3/2)² - 25/4 Min TP = (3/2,-25/4)


If graph is shifted 4 units left and 4 units up the rule becomes:
y = (x + 5/2)² - 9/4 Min TP = (-5/2, -9/4)


Expand equation:
y = (x + 5/2)² - 9/4
= (x + 5/2)(x + 5/2) - 9/4
= x² + 5x + 25/4 - 9/4
= x² + 5x + 4


Calculate y-intercept: let x = 0
y = (0)² + 5(0) + 4
= 4



Find x-intercepts: let y = 0
0 = x² + 5x + 4
0 = (x + 4)(x + 1)
x = -4 or -1


Notes:
Axis of symmetry starts 3/2 units to right of y-axis and ends up 5/2 units left
- a shift to the left of 8/2 units.
y value of TP starts 25/4 units below x-axis and ends up 9/4 units below x-axis
- a shift upwards of 16/4 units

 

[HallsofIvy]

Parabola shown cuts y-axis at -4 and x-axis at x = -1 or x = 4.
Equation: y = a(x + 1)(x -4)
We don't know the value of 'a' - the dilation factor.

What, exactly, was the given problem? There exist an infinite number of such parabolas. What additional information are you given?

We will assume a = 1 and work out the minimum y- value of the parabola to
see if it coincides with the value shown on the graph (a little less than -6).

What value? What graph? For what value of x is the y value "a little less than -6"?

y = (x + 1)(x -4)
= x² - 3x - 4
Axis of symmetry is half-way between x = -1 and x = 4.
Therefore x = 3/2 is the equation of this line.

The equation of what line? The axis of symmetry? The axis of symmetry of a parabola is always "half way between" the two x-intercepts. This is not new information so does not help identify the parabola. More importantly, what is the y value when x= 3/2?

Work out y-value of TP by substituting x = 3/2 into equation:
y = (x + 1)(x - 4)
= (3/2 + 1)(3/2 - 4)
= 5/2 * -5/2
= -25/4
Graph turns just below -6 so this tells us a = 1.

"Just below -6" is meaningless. What exact number is it? You appear to be assuming "a= 1" and then deciding "well, that's close enough".

Write rule in TP form:
y = (x - 3/2)² - 25/4 Min TP = (3/2,-25/4)

I have no idea what "MIN TP" is intended to mean. It appears to be the y coordinate of the vertex.

If graph is shifted 4 units left and 4 units up the rule becomes:
y = (x + 5/2)² - 9/4 Min TP = (-5/2, -9/4)


Expand equation:
y = (x + 5/2)² - 9/4
= (x + 5/2)(x + 5/2) - 9/4
= x² + 5x + 25/4 - 9/4
= x² + 5x + 4


Calculate y-intercept: let x = 0
y = (0)² + 5(0) + 4
= 4



Find x-intercepts: let y = 0
0 = x² + 5x + 4
0 = (x + 4)(x + 1)
x = -4 or -1


Notes:
Axis of symmetry starts 3/2 units to right of y-axis and ends up 5/2 units left
- a shift to the left of 8/2 units.
y value of TP starts 25/4 units below x-axis and ends up 9/4 units below x-axis
- a shift upwards of 16/4 units

Do you have a question?

 

[stapel]

Parabola shown cuts y-axis at -4 and x-axis at x = -1 or x = 4....

Do you have a question?

No; he just likes to post complete worked solutions (which don't always relate to the actual question, and aren't always correct, or understandable), generally to questions which have already been answered (and the student has successfully completed). :shock: