If you have more questions -- just post them or e-mail them to me. I will answer them on the blog so that all your classmates can benefit from the question!
3. Given a quadratic equation, find the value of the discriminant, state
the number and nature of the roots, and tell whether the equation can be solved
by factoring
- Remember that the discriminant is the value that "DETERMINES" what kind of solutions (roots) you will have. The discriminant is calculated using b^2 - 4ac.
- If that value is negative then the nature of the roots is that they are not real; hence we say that there are no real solutions.
- If that value is ZERO then the nature of the roots is that there will be exactly one rational solution. This means that the equation is factorable.
- If that value is positive but NOT a perfect square then the nature of the roots is that there are TWO real, irrational solutions. This means the equation is not factorable.
- If that value is positive AND a perfect square then the nature of the roots is that there are TWO real, rational solutions are not real; hence we say that there are no real solutions. This means the equation is not factorable.
4a. Given the
graph of an absolute value function, write the equation of the function
in the form y = a Ix-hI+k.
Locate the vertex of the function that is (h,k). The look at the slope of the lines as that will determine the value of a. Remember that a will be negative if the function opens down (the vertex is the maximum y-value), while a will be positive if the function opens up (the vertex is the minimum y-value).
7. Given the equation of a quadratic function in standard form, find the vertex of the function algebraically using the formula x = (-b)/2a to find the x-coordinate of the vertex and using the equation of the function to find the y-coordinate.
This stems for the quadratic formula and the completing the square work we did earlier in the unit. It works great when you have a quadratic function in standard from but "a" is not equal to one. Hence completing the square to find the vertex is a mess. Don't stress over where the formula comes from -- JUST MEMORIZE IT!! Trust me on this -- where it comes from will make sense after you use it for a while.
Practice Links from Regents Prep:
Locate the vertex of the function that is (h,k). The look at the slope of the lines as that will determine the value of a. Remember that a will be negative if the function opens down (the vertex is the maximum y-value), while a will be positive if the function opens up (the vertex is the minimum y-value).
6. Given the equation of a quadratic function in standard form, y = ax^2+bx+c where a = 1, rewrite the function in vertex
form by completing the square. State the
vertex, AOS, and range.
Remember that Vertex form looks like y = a (x-h)^2+k. So if you complete the square using the format given in class, then the vertex is just (h,k) the AOS (axis of symmetry) is x=h. The range will be based on k. If the quadratic opens up (hence k is the minimum value) then the range is {y:y>=k} (y is greater than or equal to k). If the quadratic opens down (hence k is the maximum value) then {y:y<=k) (y is less than or equal to k).7. Given the equation of a quadratic function in standard form, find the vertex of the function algebraically using the formula x = (-b)/2a to find the x-coordinate of the vertex and using the equation of the function to find the y-coordinate.
This stems for the quadratic formula and the completing the square work we did earlier in the unit. It works great when you have a quadratic function in standard from but "a" is not equal to one. Hence completing the square to find the vertex is a mess. Don't stress over where the formula comes from -- JUST MEMORIZE IT!! Trust me on this -- where it comes from will make sense after you use it for a while.
9. Given a quadratic function in standard form y=ax^2+bx+c, describe how the graph of
the function will be transformed if the value of a is changed in specified ways.
Remember if the absolute value of a is greater than 1 then the quadratic has been "stretched" -- the parabola will be narrower.
If the absolute value of a is less than 1 then the quadratic has bee compressed -- the parabola will be wider.
If a is positive the parabola will open up, while if a is negative the parabola will open down.
Practice Links from Regents Prep:
- Solving Quadratics: http://www.regentsprep.org/Regents/math/ALGEBRA/AE5/indexAE5.htm
- Solving a linear-quadratic system by graphing: http://www.regentsprep.org/Regents/math/ALGEBRA/AE6/LLinQuad1.htm
- Exponential Growth & Decay Practice: http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/indexAE7.htm
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