what is the equation in standard form of a parabola that models the values in the table x -2 0 4 f(x) 0 -6 78
Question
Answer:
Answer: [tex]y=4x^2+5x-6[/tex] Step-by-step explanation: Given points are : x = -2 0 4y=f(x) = 0 -6 78 We have to model the parabola with the help of these pointsSolution: We consider a standard equation of parabola [tex]y= ax^2+bx+c[/tex] now, we put the points in the equation we get,at (-2,0) is 4a-2b+c=0at (0,-6) is c=-6at (4,78) is 78= 16a+4b+c now, solving these equation we get, a= 4 , b= 5 , c= -6 so the equation formed with these points is [tex]y=4x^2+5x-6[/tex] we can see this in the graph attached.
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10 months ago
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