Find $y = P(x) $ for $x=4,18$

Asked Sep 14 '17 at 16:11

Active Sep 14 '17 at 16:11

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Let $P(x)= ax^2+bx+c$ whose coefficients are approximate numbers: $a= 2,51$, $b=-0,89$, $c=2,84$

Find $y= p(x) for x=4,1800000000$ and find $\delta_y$ the absolute error of the answer.

My attempt:

To find $y$ all I have to do is plug in the value. I get: $y= 42,9748$
Now to find the absolute value I plugged in for $a,b, c$ and $x$ the values that would give me the biggest result: $P(x)_{max} = 2,515 \times 4,1800000005^2 -0,885 \times 4,1799999995 + 2,845 = 43.088786011$ Now $\delta_y = |P(x)_{max} - y| = 43.088786011-42,9748 =0.113986011$

Is that the right way to approach the problem?

asked Sep 14 '17 at 16:11

To clarify: you are using the comma $,$ as a decimal point, i.e., $0.001 = 0,001$? – Sean Roberson Sep 14 '17 at 16:19
@SeanRoberson Yes! should I edit my post? – Sylvester Stallone Sep 14 '17 at 16:32
If you wish, it isn't required. – Sean Roberson Sep 14 '17 at 16:35
Are you sure that the coefficients are with errors? And would you not have to apply the same sort of error to the coefficients as all numbers use the same data type? Have you checked the error towards the smallest possible result? – Lutz Lehmann Sep 14 '17 at 17:47

0 Answers0