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f(x) = 2x^4 + 4x^3 - 7x^2 - x + 7 Use the graphing calculator to approximate the...

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kristenmarieb... | Student, Grade 10 | Valedictorian

Posted July 26, 2013 at 4:55 PM via web

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f(x) = 2x^4 + 4x^3 - 7x^2 - x + 7

Use the graphing calculator to approximate the irrational solutions correct to 3 decimals.

If there is more than 1 Real solution, enter them from smallest to largest.

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

Posted July 26, 2013 at 5:05 PM (Answer #1)

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Find the real zeros for `f(x)=2x^4+4x^3-7x^2-x+7` :

The possibilities are 4 real zeros, 2 real zeros, or no real zeros. We can look at the graph -- any x-intercepts are real zeros:

There are 3 "turning points"' this is where the graph changes from increasing to decreasing or vice versa. This is the maximum number of turning points for a quartic. For a quartic with positive leading coefficient the graph grows without bound as x tends to positive or negative infinity.

So from the graph there are 2 real zeros and 2 imaginary zeros.

Using a graphing utility we can estimate the zeros:

We see a zero between -3 and -2 which is `x~~-2.984853` and we see a zero between -1 and 0 which is `x~~-.9396253`

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To three decimal places, the values of the two real zeros are -2.985 and -0.940

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