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Mariulka [41]
2 years ago
15

The graph of a quadratic f(x) intersects the x-axis at –15 and 9. What is a possible

Mathematics
1 answer:
Orlov [11]2 years ago
5 0

the point that the function intercepts the x axis are the zeros of the function, where the products set equal to zero give you the x coordinate in which it intercepts.

x=-15 and x=9

set both equal to 0

x+15 and x-9

These are your two products

(x+15)(x-9)

or

x^2+6x-135

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Solve for x<br> Urgent need help
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I believe you need to solve this using the quadratic formula!
To begin, this is what it is:
x= -b ± <span>√ b^2 - 4ac / 2a
Just plug in what you have in your problem...
2 being a, 13 being b, and -24 being c.
So we get:
x= -13 </span>± <span>√13^2 - 4(2)(-24) / 2(2)
x= -13 </span><span>± √169 - 8 (-24) / 4</span>
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x= -13 </span>± √<span>361 / 4
The square root of 361 is 19.
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Here's where you take the equation </span>-13 <span>± 19 and put the addition and subtraction sign to use.
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Now all is left to do is divide the two numbers by 4.
-32/4 = -8
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6 0
3 years ago
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=  \frac{6x + 8}{4x + 4}  \div  \frac{4x + 4}{4x + 4}   \\  =  \frac{2x + 4}{1}  \\  = 2x + 4
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