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The Axis of Symmetry of Parabola

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College Algebra (MAT111a)

   

Added on  2020-03-07

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Question Five Question Six The graph of f(x) falls the left and falls to the right; this is because Question Seven The graph of f(x) falls the left and falls to the right; this is because Thus x intercepts are At y intercept, Hence y intercept When a graph is symmetrical about the x-axis, Thus When a graph is symmetrical about y axis, Thus Question Eight Since 0.76 is positive and the power of x, 3, is odd, the function

The Axis of Symmetry of Parabola

   

College Algebra (MAT111a)

   Added on 2020-03-07

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Question Onej(x)=−x2−1Question Twof(x)=2x2−8x+3Giventheparabolay=a(x−h)2+k,thevertexisthecoordinate(h,k)Thusthevertexoftheparabolaisgivenby−b2ax=−b2a=−−82∗2=2f(x)=2x2−8x+3=−5Thus The vertex is at x = _2____ and y = _-5____ Question Threea)f(x)=4−¿Since the equation is in the form of parabolay=a(x−h)2+k,thevertexis(h,k)=(1,4)Whenx=0,f(x)=3Whenf(x)=0,x=3,−1Thus we have the following points to plot: (1,4),(0,3),(3,0)∧(−1,0)
The Axis of Symmetry of Parabola_1
b)The axis of symmetryof aparabolais a vertical line that divides the parabolainto two congruent halves. From the graph above, equation of the parabola’s axis of symmetry is:x=1c)Domain of the function is given by: −∞<x<∞The range of the function is given by: −∞<f(x)≤4Question Foura)Themaximumheight∧thedistance¿thepointitwasthrownwillbegivenbyvertexoftheparabolaf(x)=−0.8x2+2.4x+6Atthevertex,x=−b2a=−2.42∗−0.8=1.5Whenx=1.5,f(x)=7.8Thus the ball’s maximum height ¿7.8feet∧thedistance¿whereitwasthrownhorizontally=1.5feet
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b)Atthepointtheballhitstheground,f(x)=0Implying−0.8x2+2.4x+6=0Usingquadraticformula,x=3+√392≈4.6225Thus the ball travels 4.6 feet horizontally before hitting the ground. c)Question Fivef(x)=7x5+2x2+1xQuestion SixThe graph of f(x) falls the left and falls to the right; this is because−5isnegativewhilethepower,4,iseven.Question Seven
The Axis of Symmetry of Parabola_3

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