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Ann [662]
3 years ago
7

Write a quadratic function with a graph that meets the following conditions: x-intercepts at (2, 0) and (6, 0) with a minimum at

(4, -8)
Mathematics
1 answer:
Darya [45]3 years ago
5 0
The minimum is the vertex
y=a(x-h)^2+k
(h,k) is vertex
given
(4,-8)
y=a(x-4)^2-8
find a
given
(2,0) and (6,0)
find a

0=a(2-4)^2-8
0=a(2)^2-8
0=4a-8
8=4a
divide 4
2=a

other one
0=a(6-4)^2-8
0=a(2)^2-8
0=4a-8
8=4a
divide 4
2=a


the function is
y=2(x-4)²-8 or expanded
y=2x^2-16x+24
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Answer:

D) x=\frac{-2+-2\sqrt{7} }{3}

Step-by-step explanation:

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A in terms of this question=9

B in terms of the question is 12

C in terms of the question is -24.

This question is an example of a quadratic equation. To work this out you may first need a calculator. The first step is to substitute the values of a,b and c into the formula.  So once substituted the formula of \frac{-b+-\sqrt{b^2-4ac} }{2a} becomes \frac{(-12)+-\sqrt{(12)^2-4*(9)*(-24)} }{2(9)}. Although when written in a calculator there will not be a plus and minus button and so you would have to do this separately.

However when substituting the values it would be best practice to put them in brackets.

1) Substitute the values into the equation for +.

\frac{(-12)+\sqrt{(12)^2-4*(9)*(-24)} }{2(9)}=\frac{-2+2\sqrt{7} }{3} /1.097167541

2) Substitute the values into the equation for -.

\frac{(-12)-\sqrt{(12)^2-4*(9)*(-24)} }{2(9)}=\frac{-2+2\sqrt{7} }{3} /-2.430500874

8 0
3 years ago
What is the solution to the system of equations?
Tatiana [17]
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y = –5x + 3
<span>y = 1
</span>_______________
1 =<span> –5x + 3</span> 
5x = 3 - 1
5x = 2    /5
x = 2/5 = 0.4

(x, y) = (0.4, 1)

The correct result would be (0.4, 1<span>).</span>
3 0
3 years ago
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