Download Honors PreCalculus Chapter P Tips (help with test corrections) 1) To

Honors PreCalculus Chapter P Tips
(help with test corrections)
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To find the slope and y-intercept of a linear equation, put the linear equation in slope-intercept form (y =
mx + b). Where “m” is the slope and “b” is the y-intercept.
When solving a radical equation you must first isolate the radical part (when it contains the variable) by
moving everything else on the other side of the equal sign. Then, square both sides to eliminate the square
root. Solve for the variable!
When solving rational equations, clear out the fractions! Determine the LCD every term in the equation
and multiply every term by the LCD. You’ll notice the denominator in each term will cancel out with the
part of the LCD it matches. Multiply the numerator by the “left-over parts” of the LCD. Distribute,
combine like terms, and isolate the variable you are trying to solve for. You must check your answer to be
sure there aren’t any restrictions on the variable. Check by plugging your answer in for the variable in the
original equation.
When two lines are parallel they have the same slope. When two lines are perpendicular they have
opposite reciprocal slopes. In order to find an equation perpendicular or parallel to a given line and goes
through a given point, you must first determine the slope of the line you are given. Once you know the
slope you need to substitute the necessary slope (the same if parallel, opposite reciprocal if perpendicular)
into the slope-intercept form of a line with the ordered pair you were given. Solve for “b”. After finding b,
plug back into the slope-intercept form of the equation. NOTE: You can also use the point-slope form!
When points are collinear the slope is the same between any given points. Determine the slope of two
points you are given. Once you have the slope, set it equal to the unknown point and one of the points you
are given. Cross multiply to solve for the unknown.
The diameter of a circle is the length from one point on the circle to another. The diameter goes through
the center of the circle. The center of the circle is the midpoint of the diameter! Use the midpoint formula
to find the center of the circle. Once you have the center of the circle you can find the length of the radius
by using the distance formula. You now have everything you need in order to come up with the equation of
a circle. The center is (h, k) and the radius is r. Just plug it in!
When trying to solve a polynomial inequality there is a series of steps you must follow. There are
wonderful examples of problems like this in the book (section P. 5). The first thing you need to do is get
all terms on one side leaving “0” on one side. Factor the polynomial, if possible. Set each factor equal to
zero to find your critical points. If the equation is a quadratic, there will be two critical points! The critical
points determine the intervals in which to test sign changes on the graph. If there are two critical points,
there will be three test intervals. Pick a point in each test interval to see if its result will be positive or
negative. Look back at your inequality. Use the direction of the inequality symbol to see whether or not
you want positive or negative results.
Vertex form of a quadratic is very similar to completing the square to a quadratic. Again, a series of steps
need to be taken. First: put parentheses around the QT and LT of the quadratic. If the coefficient of the
QT is something other than “1”, divide the QT and the LT by that number (it goes on the outside of the
parentheses!). Second: take ½ of the LT coefficient and square it. Add the result of the second step to the
QT and LT in the parentheses. Since you changed the integrity of the equation you have to “undo”it by
taking the number off the back of the equation by subtracting it. Remember, if you divided by a number
(the number on the outside of the parentheses) you need to multiply it by the number you added and
subtract that result from the back of the equation. Now what you have in the parentheses is a perfect square
trinomial- it’s factorable!!! Write it as a binomial squared and you are in vertex form!
Absolute value equations always look like a “V” when graphed. The standard form of an absolute value
equation looks just like the vertex form of a quadratic. The (h, k) is the vertex, the number in front of the
absolute value symbols determine the direction of the opening, h shows how it shifts left or right, and k
determines how it shifts up or down. First graph the vertex and choose an x value on either side of the
vertex to determine the y value that goes with it.
Given a line segment, the midpoint is the “middle point”- exactly ½ way in between the two endpoints. If
you are given one endpoint and the midpoint, use the midpoint formula to find the other endpoint. Drawing
a picture will help tremendously!
11) The zero’s of an equation is the location in which the graph crosses the x-axis. In order to find the zero’s of
a quadratic you can factor, complete the square, or quadratic formula. Just plug the necessary coefficients
into the formula and find your zero’s!
12) Finding x and y intercepts of any equation involves the same steps. If you are trying to find the xintercept(s), set y = 0 and solve for x. This may require you to factor! To find they-intercept(s), set x = 0
and solve for y. It’s that easy!
***Note: You have a lot of tools in your toolbox. Your notes, your quizzes, your textbook. Use them! Search
through what you have to get examples on how to do these problems.
These concepts in Chapter P are all review. You’ve seen them in Algebra 2, Algebra I, and/or Geometry. Whether
you learned them or not, you must learn them now! These concepts are imperative for Honors Pre-Calculus and you
must need to do them with ease. As the material gets more difficult you cannot afford to get hung up on these basic
concepts from previous classes.
Yes, I am here to help you however you need to try and help yourself first!
Chapter P is over…we’re moving on. Get ready, be ready.