Find a quadratic function with the given x-values as the only two REAL zeros. x=-2 and x=8. Please show work! Will mark Brainlie
1 answer:
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Answer:
y = 4x² -3x +5
Step-by-step explanation:
It can be helpful to plot the points on a graph. The curve through them opens upward, so you know <em>the first two answer choices are incorrect</em>.
<em>Looking at the location of the vertex</em>
The points (-1, 12) and (2, 15) have nearly the same y-value, so the vertex of the parabola will be nearly halfway between those points, at about ...
... x = (-1 +2)/2 = 1/2
Because the point at (2, 15) has a slightly higher y-value, you know it is farther from the vertex, so the vertex will be at an x-value that is less than 1/2.
For quadratic ...
... y = ax² +bx +c
the vertex is found at
... x = -b/(2a)
For the 3rd choice, this is x = -(-4)/(2·3) = 2/3. This is higher than 1/2, so we can eliminate this choice, too.
Only the 4th choice remains. (It passes the vertex location test: x = -(-3)/(2·4) = 3/8 < 1/2.)
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There are some other ways to solve this problem, too.
- try the given points in the given equations. The +5 as a constant of all of them tell you that all will work for x=0. Both of choices 3 and 4 work for x=-1, so we have to try x=2. Choice 3: 3·2² -4·2 +5 = 9, not 15. Choice 4: 4·2² -3·2 +5 = 15. So Choice 4 is the answer.
- write equations for a, b and solve them. a(-1)² +b(-1) +5=12; a(2)² +b(2)+5=15. or a-b=7, 2a+b=5 in standard form. Adding these gives 3a=12, so a=4 and Choice 4 is the answer.
- use a graphing calculator to write the quadratic regression formula from the three points. This result is shown in the attachment. The (a, b, c) = (4, -3, 5) correspond to Choice 4.
