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# I want to find the intercepting point on a linear and quadratic equation

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I want to find the intercepting point on a linear and quadratic equation

The equations are

y=2.6x

y=0.5558x2-2.482x+4.4573

The answers i got were x = 4.534 and x = 0.98

y= 2.548 and y = 11.7884

2.548 and 0.98 are both right but the higher number on the quadratics are wrong i know because i graphed it.

It should be x=8.161 and y = 21.218..........x= 0.98 y = 2.548

Guest Jun 12, 2017

#1
+90023
+1

y=2.6x

y=0.5558x^2-2.482x+4.4573

Set the y's equal.....and we have

2.6x  =  0.5558x^2-2.482x+4.4573     subtract  2.6x from borh sides

0.5558x^2 - 5.082x +4.4573  =  0

And the solutions are

x ≈  0.982689      and   x  ≈   8.16089

CPhill  Jun 12, 2017
#1
+90023
+1

y=2.6x

y=0.5558x^2-2.482x+4.4573

Set the y's equal.....and we have

2.6x  =  0.5558x^2-2.482x+4.4573     subtract  2.6x from borh sides

0.5558x^2 - 5.082x +4.4573  =  0

And the solutions are

x ≈  0.982689      and   x  ≈   8.16089

CPhill  Jun 12, 2017