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Introductory Summary of the Parabola Any Second Degree Single Variable

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Introductory Summary of the Parabola Any Second Degree Single Variable Introductory Summary of the Parabola Any second degree single variable polynomial equation is called a “quadratic equation”. In other words, if you have an equation that can be written “y =” and uses only one letter among its terms, has only whole numbers as exponents, and the largest of the exponents is the number “2” then the equation is a quadratic equation. Here are some examples of various forms quadratic equations: y=x2−4x+7 y=−2x2+6x+2 y=x2−9 x2+y−6x+7=0 Important Points 1. The graph of a quadratic equation is a U-shape and is called a parabola. Opens up Opens down Opens left Opens right 2. Where the U turns (at the bottom, the top, the left, or at the right) is called the vertex. The vertex is a point and has coordinates x and y written as an ordered pair ( x, y ). 3. There are two standard forms associated with a parabola: a). Standard Parabolic Form: y=a(x−h)2+k If you see (x+h)2, then you have to rewrite it as (x−−h)2to be in pure standard form. Likewise, if you see a negative k, y=a(x−h)2−k, then you have to rewrite the k as y=a(x−h)2+(−k) (1). The parabola opens up if “a” is positive and down if “a” is negative. (2). The vertex point is (h,k) where “h” is to the right of the minus sign inside the parentheses raised to the power of 2. Be sure to include the sign of the number. In (x−−4)2, h is negative 4. “k” is to the right of the plus sign outside the parentheses. In y=5(x−3)2−7=5(x−3)2+(−7), the “k” is negative 7. b). Standard Quadratic Form: y=ax2+bx+c In this form, (1). The parabola opens up if “a” is positive, and opens down if “a” is negative. (2). The vertex is a point (h,k). You have to do some calculations to find h and k. −b h= 2a Once you have a value for h, then go back to the original quadratic equation y=ax2+bx+c and replace each “x” with your “h” value, do the arithmetic, and you’ll have “k”. For example, in y=x2+4x+1 −4−4 h===−2 2(1)2 To find “k” go back to y=x2+4x+1 and replace x by −2 (−2)2+4(−2)+1 =4−8+1 =−4+1 =−3 The vertex is (−2,−3) Using a TI82, 83, 85, 86 calculator, you could calculate “x” , use the STO> key, and type the original equation to determine k. For example, we found x=−2; this is h. On the calculator I would type −2 STO>x and press the ENTER key. Then I would type x^2+4x+1 and press ENTER. The calculator would display −3 which is k. (3). To convert from y=ax2+bx+c form to y=a(x−h)2+k, complete the square. When you get near the end of completing the square, you may need to multiply ALL TERMS on BOTH SIDES of your equation by the number under “y”. This will produce the form y=a(x−h)2+k. (4). To find a “y-intercept”, let x=0 and solve for y. Answer (0,y). To find an “x-intercept”, let y=0 and solve the quadratic equation using any method you’ve learned. For example: extract a root, factor, complete the square, use the quadratic equation. Answer (x,0). You may get 1 real x, 2 real x values, or 2 imaginary x-values in which case there are no x-intercepts. 4. To write the equation of a parabola when given the vertex (h,k)and a point (x,y) on the parabola, follow the following steps: a). Use the form y=a(x−h)2+k b). Replace x and yby the coordinates found in the point (x,y). c). Replace h and k by the values found in the vertex (h,k). d). Do the arithmetic and solve for “a”. e). In the original equation (see step 1), replace a,h,k by the numbers you found for those letters BUT leave x and y as variables (letters). f). To write the answer in y=ax2+bx+c form, (1). Expand (x−h)2 by writing (x−h)(x−h) and multiplying. (2). Distribute the “a” that was in front of (x−h)2 to all of the terms obtained when you multiplied. (3). Add “k” to the result and simplify. .
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