2.01
CHAPTER 2:
POLYNOMIAL AND RATIONAL FUNCTIONS
SECTION 2.1: QUADRATIC FUNCTIONS (AND PARABOLAS)
PART A: BASICSIf a, b, and care real numbers, then the graph offx( )=y=ax2+bx+cis a parabola, provided a≠0If a>0, it opens upwardIf a<0, it opens downward
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2.02
PART B: FINDING THE VERTEX AND THE AXIS OF SYMMETRY (METHOD 1)The vertex of the parabola [with equation] y=ax2+bx+cis h,k(), where:xcoordinate=h=−b2a, andycoordinate=k=fh( ).The axis of symmetry, which is the vertical line containing the vertex,has equation x=h(Does the formula for hlook familiar? We will discuss this later.)ExampleFind the vertex of the parabola y=x2−6x+5. What is its axis of symmetry?
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2.03
PART C: FINDING MORE POINTS“Same” Example: y=x2−6x+5, or fx( )=x2−6x+Find the yintercept.Plug in 0 for x. Solve for yIn other words, find f0( )The yintercept is 5(If the parabola is given by y=ax2+bx+c, then c, the constant term, is theyintercept. Remember that bwas the yintercept for the line given byy=mx+b.)Find the xintercept(s), if any.
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