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f(x) = -0.1x^2 + 3.2x - 3.5 Find the x intercepts of the parabola. Thanks!

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loishy | Student, Grade 10 | Salutatorian

Posted April 29, 2013 at 9:29 PM via web

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f(x) = -0.1x^2 + 3.2x - 3.5

Find the x intercepts of the parabola. Thanks!

Tagged with math, parabola, x intercepts, x-axis

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durbanville | High School Teacher | (Level 1) Educator Emeritus

Posted April 29, 2013 at 10:01 PM (Answer #1)

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On a parabola, the x intercepts exist, of course, along the x-axis. this means that, at these points y=0. Therefore, solve the equation where y=0 and hence:

`-0.1x^2+3.2x - 3.5 = 0`  

To make it easier to work with multiply by -1 and note the symbol changes :

`0.1x^2 - 3.2x +3.5 =0`

It may be easier to work with whole numbers rather than decimal fractions and, if preferred, multiply the equation by 10:

`x^2 - 32x +35 = 0`  

Use the quadratic formula `(-b+- sqrt(b^2-4ac))/(2a)` 

where a=1;b=-32; c=35

Substituting:

`x =(-(-32) +-sqrt((32^2)- 4(1)(35)))/(2(1))`

`therefore x= (32+-sqrt(1024-140))/2`

`therefore x= (32 +sqrt884)/2`     OR    `x=(32-sqrt884)/2`

`therefore x = 30.866`        OR   `x=1.134`

therefore the x-intercepts are (30.866;0)  and (1.134; 0)

 

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oldnick | Valedictorian

Posted April 29, 2013 at 11:44 PM (Answer #2)

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`-0.1x^2+3.2x-3.5=0`

`x^2-32x+35=0`

`x^2-32x+256+35=256`

`(x-16)^2+35=256`

`(x-16)^2=221`

`x-16=+-sqrt(221)`

`x=16+-sqrt(221)`

`x_1= 30.86606874732`

`x_2=1.133931252681`

 

 

 

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