Imperial Valley College Math Lab
QUADRATICS SUMMARY SHEET (EQUATIONS, FUNCTIONS, AND GRAPHS)
The two basic forms of quadratic equations are: 푦 = 푎푥2 + 푏푥 + 푐 푦 = 푎(푥 − ℎ)2 + 푘 The two basic forms of quadratic functions are: 푓(푥) = 푎푥2 + 푏푥 + 푐 푓(푥) = 푎(푥 − ℎ)2 + 푘
SQUARE ROOT PROPERTY
If 푋2 = 푘 , then 푋 = ±√푘 . [ can be any algebraic expression, and 푘 is any real number].
Ex: (푥 − 4)2 = 5 → 푥 − 4 = ±√5 → 푥 = 4 ± √5
QUADRATIC FORMULA
−푏 ± √푏2−4푎푐 If 푎푥2 + 푏푥 + 푐 = 0 , then 푥 = . 2푎
Remark: Notice that the equation must be set equal to zero first!
Ex: 4푥2 + 5푥 − 3 = 0 First note that 푎 = 4 , 푏 = 5, and 푐 = −3 .
−5 ± √52−4(4)(−3) −5 ± √25−(−48) −5 ± 73 Substitute carefully: 푥 = = = √ . 2(4) 8 8
Watch the signs! Especially when a or c is negative.
THE DISCRIMINANT
The type of solution to a quadratic equation depends upon what happens under the radical sign (√ ) . Therefore, if you look at the Quadratic Formula above, you’ll see that the “푏2 − 4푎푐 ” part is important.
Definition: For the quadratic equation 푎푥2 + 푏푥 + 푐 = 0 , the discriminant is 푏2 − 4푎푐 .
If 푏2 − 4푎푐 > 0 and is a perfect square, then 푎푥2 + 푏푥 + 푐 = 0 has two rational solutions. If 푏2 − 4푎푐 > 0 but is not a perfect square, then 푎푥2 + 푏푥 + 푐 = 0 has two irrational solutions. If 푏2 − 4푎푐 = 0 , then 푎푥2 + 푏푥 + 푐 = 0 has one rational solution. If 푏2 − 4푎푐 < 0 , then 푎푥2 + 푏푥 + 푐 = 0 has two complex solutions (involving the imaginary number ).
*For our purposes here, irrational solutions involve square roots that are real but cannot be simplified, such as √3 or √11 , while rational solutions no longer have square root signs after simplification, e.g. √9 = 3 . The example above has two irrational solutions because of the ±√73 . GRAPHING QUADRATIC EQUATIONS
The graphing method depends on whether you have the form 푓(푥) = 푎푥2 + 푏푥 + 푐 or 푓(푥) = 푎(푥 − ℎ)2 + 푘. If 푎 ≠ 0 , the graph will always be a parabola. In both cases…
1) If 푎 > 0 , then parabola opens upward. If 푎 < 0 , the parabola opens downward.
2) Find the vertex. a. For 푓(푥) = 푎(푥 − ℎ)2 + 푘 , the vertex is (ℎ, 푘). 푏 푏 b. For 푓(푥) = 푎푥2 + 푏푥 + 푐 , the vertex is (− , 푓 (− )) . 2푎 2푎
3) Find and plot the y-intercept by setting 푥 = 0 and computing 푓(푥) .
4) Find and plot the x-intercept(s) by setting 푓(푥) = 0 (i.e. setting 푦 = 0) and finding the x-value(s). ======Example: Graph 푓(푥) = −(푥 + 1)2 + 4 and 푓(푥) = −푥2 − 2푥 + 3 : [You should verify that these are actually two different forms of the same function] [Please see a tutor for guidance through this example if necessary] 1) Since 푎 = −1 is negative, the parabola will open downward.
2) a. The vertex (ℎ, 푘) = (−1, 4) . Why does ℎ = −1? Notice in the general formula, the h must follow the minus sign. So, think of (푥 + 1)2 as (푥 − −1)2 . This means ℎ = −1 .
−푏 −(−2) 2 b. ℎ = = = = −1 (just as it did above), and 2푎 2(−1) −2 −푏 푘 = 푓 ( ) = 푓(−1) = −(−1)2 − 2(−1) + 3 = −1 + 2 + 3 = 4 (just as it did above). 2푎
3) Substituting 푥 = 0 , we obtain 푓(0) = 3 . So, the y-intercept is (0, 3) .
4) Set 푓(푥) = 0 and solve for 푥 . For practice, check both cases:
0 = −(푥 + 1)2 + 4 and 0 = −푥2 − 2푥 + 3
In both cases, you should get 푥 = −3 and 푥 = 1 . So the x-intercepts are (−3, 0) and (1, 0) . y
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