All SAT II Math I Resources
Example Questions
Example Question #2 : Quadratic Inequalities
Solve:
Start by changing the less than sign to an equal sign and solve for .
Now, plot these two numbers on a number line.
Notice how the number line is divided into three regions:
Now, choose a number fromeach of these regions to plug back into the inequality to test if the inequality holds.
For , let
Since this number is not less than zero, the solution cannot be found in this region.
For , let
Since this number is less than zero, the solution can be found in this region.
For let .
Since this number is not less than zero, the solution cannot be found in this region.
Because the solution is only negative in the interval , that must be the solution.
Example Question #10 : Quadratic Inequalities
Which value for would satisfy the inequality ?
Not enough information to solve
First, we can factor the quadratic to give us a better understanding of its graph. Factoring gives us: . Now we know that the quadratic has zeros at and . Furthermore this information reveals that the quadratic is positive. Using this information, we can sketch a graph like this:
We can see that the parabola is below the x-axis (in other words, less than ) between these two zeros and .
The only x-value satisfying the inequality is .
The value of would work if the inequality were inclusive, but since it is strictly less than instead of less than or equal to , that value will not work.
Example Question #1 : Solving Other Functions
Simplify:
You may assume that is a nonnegative real number.
The best way to simplify a radical within a radical is to rewrite each root as a fractional exponent, then convert back.
First, rewrite the roots as exponents.
Multiply the exponents, per the power of a power rule:
Example Question #1 : Solving Other Functions
Define functions and .
for exactly one value of on the interval .
Which of the following statements is correct about ?
Define
Then if ,
it follows that
,
or, equivalently,
.
By the Intermediate Value Theorem (IVT), if is a continuous function, and and are of unlike sign, then for some . As a polynomial, is a continuous function, so the IVT applies here.
Evaluate for each of the following values: :
Only in the case of does it hold that assumes a different sign at both endpoints - . By the IVT, , and , for some .
Example Question #1 : Graphing Linear Functions
Refer to the above red line. A line is drawn perpendicular to that line, and with the same -intercept. Give the equation of that line in slope-intercept form.
First, we need to find the slope of the above line.
The slope of a line. given two points can be calculated using the slope formula
Set :
The slope of a line perpendicular to it has as its slope the opposite of the reciprocal of 2, which would be . Since we want this line to have the same -intercept as the first line, which is the point , we can substitute and in the slope-intercept form:
Example Question #402 : Sat Subject Test In Math I
Refer to the above diagram. If the red line passes through the point , what is the value of ?
One way to answer this is to first find the equation of the line.
The slope of a line. given two points can be calculated using the slope formula
Set :
The line has slope 3 and -intercept , so we can substitute in the slope-intercept form:
Now substitute 4 for and for and solve for :
Example Question #111 : Functions And Graphs
Which equation best matches the graph of the line shown above?
To find an equation of a line, we will always need to know the slope of that line -- and to find the slope, we need at least two points. It looks like we have (0, -3) and (12,0), which we'll call point 1 and point 2, respectively.
Now we need to plug in a point on the line into an equation for a line. We can use either slope-intercept form or point-slope form, but since the answer choices are in point-slope form, let's use that.
Unfortunately, that's not one of the answer choices. That's because we didn't pick the same point to substitute into our equation as the answer choices did. But we can see if any of the answer choices are equivalent to what we found. Our equation is equal to:
which is the slope-intercept form of the line. We have to put all the other answer choices into slope-intercept to see if they match. The only one that works is this one:
Example Question #2 : Graphing Linear Functions
Determine where the graphs of the following equations will intersect.
We can solve the system of equations using the substitution method.
Solve for in the second equation.
Substitute this value of into the first equation.
Now we can solve for .
Solve for using the first equation with this new value of .
The solution is the ordered pair .
Example Question #1 : Graphing Linear Functions
Refer to the line in the above diagram. It we were to continue to draw it so that it intersects the -axis, where would its -intercept be?
First, we need to find the slope of the line.
In order to move from the lower left point to the upper right point, it is necessary to move up five units and right three units. This is a rise of 5 and a run of 3. makes the slope of the line shown .
We can use this to find the -intercept using the slope formula as follows:
The lower left point has coordinates . Therefore, we can set up and solve for in this slope formula, setting :
Example Question #1 : Graphing Linear Functions
Line includes the points and . Line includes the points and . Which of the following statements is true of these lines?
The lines are distinct but neither parallel nor perpendicular.
The lines are parallel.
The lines are perpendicular.
Insufficient information is given to answer this question.
The lines are identical.
The lines are parallel.
We calculate the slopes of the lines using the slope formula.
The slope of line is
The slope of line is
The lines have the same slope, making them either parallel or identical.
Since the slope of each line is 0, both lines are horizontal, and the equation of each takes the form , where is the -coordinate of each point on the line. Therefore, line and line have equations and .This makes them parallel lines.
Certified Tutor