All SAT II Math I Resources
Example Questions
Example Question #1 : Transformations
Reflect the point across the line
and then across the origin.
Reflecting the point across the vertical line will only change the x-value, but not the y-value.
The point after this reflection is:
Rotating this point across the origin will swap the x and y-values.
The new point is:
Example Question #1 : How To Find Transformation For An Analytic Geometry Equation
Let . If
is equal to
when flipped across the x-axis, what is the equation for
?
When a function is flipped across the x-axis, the new function
is equal to
. Therefore, our function
is equal to:
Our final answer is therefore
Example Question #1 : Transformation
Let . If we let
equal
when it is flipped across the y-axis, what is the equation for
?
When a function is flipped across the y-axis, the resulting function
is equal to
. Therefore, to find our
, we must substitute in
for every
is our equation:
Our final answer is therefore
Example Question #1 : Transformation
Let . If
represents
is shifted
places to the right and
places upwards, what is the equation for
?
When a function is transformed
units upwards, the new function
is equal to
. Likewise, if
is transformed
units to the right, the new function
is equal to
. Therfore, we can first find the upwards transformation by adding
to the function:
Now we can apply the horizontal transformation by replacing all 's in the function with
. Our transformed function therefore becomes:
We then multiply this out to obtain:
Our final answer is therefore
Example Question #1 : Transformations
How is different from
?
The graph of is dilated compared to the graph of
.
The graph of is shifted to the right three units along the x-axis from the graph of
.
The slope of is steeper than the slope of
.
The graph of is shifted up three units along the y-axis from the graph of
.
The slope of is steeper than the slope of
.
The standard form of a linear equation is Here, we are given two equations,
and
, which differ only in their
terms. In other words, these functions differ only in their slope.
has a larger slope than does
, so
is steeper.
Example Question #12 : Coordinate Geometry
Given , write an equation
that represents a vertical shift four units upward.
Algebraic transformations of functions rely on manipulating components of the equation's standard form. The standard form of a linear equation is . Changes to the slope (
) will make the graph steeper or shallower, changes to the y-intercept (
) will shift the graph vertically, and changes to the indepent variable (
) will shift the graph horizontally. Here, we are given
and asked to transform it into a new equation vertically shifted up four units. We can accomplish this by adding four to the constant term, so the correct answer is
.
multiples the slope by four, which will result in a steeper graph.
subtracts four from the constant term, which shifts the graph vertically, but in the wrong direction.
adds four to the independent variable, which shifts the graph horizontally to the left.
Example Question #4 : How To Find Transformation For An Analytic Geometry Equation
Given , write an equation
that represents a horizontal shift two units to the right.
Algebraic transformations of functions rely on manipulating components of the equation's standard form. The standard form of a linear equation is . Changes to the slope (
) will make the graph steeper or shallower, changes to the y-intercept (
) will shift the graph vertically, and changes to the indepent variable (
) will shift the graph horizontally. Here, we are given
and asked to transform it into a new equation horizontally shifted to the right two units. We can accomplish this by subtracting two from the independent variable, so the correct answer is
.
adds two to the constant term, which shifts the graph vertically.
adds two to the independent variable which shifts the graph horizontally, but in the wrong direction.
multiplies the slope by two, which makes the graph steeper.
Example Question #1 : Transformation
Given , write an equation
that increases the slope by three, shifts the graph horizontally one unit to the left, and shifts the graph vertically three units down.
Algebraic transformations of functions rely on manipulating components of the equation's standard form. The standard form of a linear equation is . Changes to the slope (
) will make the graph steeper or shallower, changes to the y-intercept (
) will shift the graph vertically, and changes to the indepent variable (
) will shift the graph horizontally. Here, we are given
and asked to transform it into a new equation that increases the slope by three, shifts the graph horizontally one unit to the left, and shifts the graph vertically three units down.
First, multiply the slope by three.
Add one to the independent variable.
Subtract three from the constant term.
correctly increases the slope and subtracts three from the constant term, but fails to properly substitute
for
, leading to an erroneous simplification. In other words,
becomes
.
shifts the function to the right instead of the left.
shifts the function both right and up, rather than left and down.
Example Question #1 : How To Find Transformation For An Analytic Geometry Equation
If this is a sine graph, what is the phase displacement?
π
0
2π
4π
(1/2)π
0
The phase displacement is the shift from the center of the graph. Since this is a sine graph and the sin(0) = 0, this is in phase.
Example Question #11 : Coordinate Geometry
If this is a cosine graph, what is the phase displacement?
(1/2)π
0
2π
π
4π
π
The phase displacement is the shift of the graph. Since cos(0) = 1, the phase shift is π because the graph is at its high point then.
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