Categorii: Tot - equations - factoring

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Quadratics Functions

Quadratic functions are a fundamental concept in algebra, involving equations of the form ax^2 + bx + c = 0. Various methods exist to solve these equations, including factoring, using the FOIL method, completing the square, and applying the quadratic formula.

Quadratics Functions

Quadratics Functions

Parts of Parabolas and Definitions

Zeros/Roots --------------- When the parabola crosses the x-axis, the x-coordinate it is called zeros or the x-intercept.
Axis of Symmetry (AOS) ---------------------------- A vertical line that depicts the point of symmetry
Vertex --------------------- A parabola has a minimum (opens upward) & maximum (opens downward) value. It is also the point of AOS.
Optimal Value ---------------- The y-value of the vertex.
Y-Intercept -------------- The coordinate where the parabola crosses the y-axis.

Solving Quadratic Equations

Using Quadratic Formula
Discriminant ------------- b^2 - 4ac 1) If the discriminant is greater than 0, then there are 2 real roots 2)If the discriminant is equal to 0, then there is 1 real root. 3) If the discriminant is than 0, then there are no real roots.
Use this formula when you cannot factor.
Complete the Square
Convert Standard Form into Vertex Form and use this to find min./max. value or vertex of parabola.

1) Start by factoring out the a 2) Move the c term to the other side of the equation. 3) Use the b term in order to find a new c term that makes a perfect square. This is done by first dividing the b term by 2 and squaring the quotient and add to both sides of the equation. 4) Find your h, the b term divided by two, for the perfect square. 5) Set equation to zero.

x^2+2x-8=0 (x+1)^2 - (1)^2 - 8 =0 (x+1)^2-1-8=0 (x+1)^2-9=0 (x+1)^2=9 x+1=+- (square root of 9) x=-3-1 x=3-1 x=-4 x=2

Expanding
FOIL Method --------------- First Outer Inner Last

(x+5)(x+2) =x^2+2x+5x+10 =x^2+7x+10

Factor
Difference of Squares -------------------------- (a+b)(a-b) = a^2 - b^2

(9x^2-16) =(3x)^2 - 4^2 = (3x+4)(3x-4)

Perfect Square Trinomial ----------------------- (a+b)^2 = a^2 + 2ab + b^2 (a-b)^2 = a^2 - 2ab - b^2

(3x+2y)^2 =3^2 + 2(3)(2) + (2)^2 = 9x^2+12xy+4y^2

Complex Trinomial
Simple Trinomial

Find 2 numbers that multiply to c and add to b.

1) Use Criss-Cross Method 2) MAN Method (Multiply, Add, Number)

x^2-2x-15 M- -15 A- -2 N- (-5,3) =(x-5)(x+3)

Common Factor

Factor by Grouping ----------------------- 1) group terms with like terms 2)factor each group to get a binomial common factor

3x^2+6x+4x+8 =(3x^2+6x)+(4x+8) = 3x(x+2)+4(x+2) = (3x+4)(x+2)

Find the GCF and take it out by dividing each term by the GCF.

6x^2 - 2x = 0 2x(3x-1)=0 2x=0 3x-1=0 x=0 x=1/3

Forms of Quadratic Equations

Factored Form: y=a(x-r)(x-s)
To find the x-value of the vertex, use the formula (r+s)/2
Binomials (x-r) and (x-s) gives the x - intercepts.

y= 0.5(x-6)(x+2) -------------------- (x-6)= 0 (x+2)=0 x=0+6 x=0-2 x= 6 x= -2

Standard Form: y=ax^2+bx+c
A rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground t seconds after it is thrown is given by d(t) = -16t^2 – 4t + 382. How long after the rock is thrown is it 370 feet from the ground? --------------------------------------------------- d(t) = 16t^2 - 4t +382 -16t^2 - 4t + 382 = 370 -16t^2 - 4t + 382 - 370 = 0 -16t^2 - 4t + 12 = 0 16t^2 + 4t - 12 = 0 4t^2 + t - 3 = 0 (4t - 3)(t + 1) = 0 4t - 3 = 0 or t + 1 = 0 t = 3/4 or t = -1 Therefore, it takes 75 seconds to reach 370 feet.
a,b, and c are real numbers and a is not equal to 0
Formula for AOS x = -b/2a
a is the stretch/ compression factor

Step Pattern --------------------- Multiply a value and step pattern to get the correct points of the parabola (1,3,5) * a

Transformation of Parabola -------------------------------- If a>1 then the graph is stretching vertically by a value. If 0

Direction of Opening ------------------------- If a>0 it is opening upwards. If a<0 it is opening downwards.

c value is the y-intercept of the parabola
Vertex Form: y=a(x-h)^2+k
Ming throws a stone off a bridge into a river below. The stone's height (in meters above the water), x seconds after Ming threw it, it is modeled by: h(x)=-5(x-1)^2+45 What is the maximum height that the stone will reach? The maximum height is reached at the vertex. So, in order to find the maximum height, we need to find the vertex's y-coordinate. The vertex is (1, 45). So, in conclusion, the stone reaches the maximum height at 45 meters.
Transformation -------------------------- y= (x - h)^2, then the parabola is shifted right by h units. y= (x+ h)^2, then the parabola is shifted left by h units y= x^2 - k, then the parabola is shifted down by k units. y= x^2 + k, then the parabola is shifted up by k units.
Vertex/Axis Of Symmetry (AOS) = (h,k)

k is the y value of vertex and AOS

If k>0 is a vertical shift up by k units. If k<0 is a vertical shift down by k units.

h is the x value of vertex and AOS

If h>0 is a horizontal shift to the right. If h<0 is a horizontal shift to the left.