URGENT WILL GIVE BRAINLIEST
2 answers:
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In this we know all three zeros and one point from which the graph pass.
So we will let specific cubic polynomial function of the form
As we know zeros are that point where we will get value of function equal to zero. So it is basically in form
SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say
So required equation is
![= a[(x^2 - 2x - 3x + 6)(x-5)]](https://tex.z-dn.net/?f=%3D%20a%5B%28x%5E2%20-%202x%20-%203x%20%2B%206%29%28x-5%29%5D)
![= a[(x^2 - 5x+6)(x-5)]](https://tex.z-dn.net/?f=%3D%20a%5B%28x%5E2%20-%205x%2B6%29%28x-5%29%5D)
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
So required equation of cubic polynomial is
For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.
So we will let specific cubic polynomial function of the form
As we know zeros are that point where we will get value of function equal to zero. So it is basically in form
SO in given question zeros are (2 , 0) , (3, 0) and (5,0)
So we can say
So required equation is
Now we have one point (0 , -5) from which graph passes.
So we say at x = 0 , f(x) = -5
So required equation of cubic polynomial is
For finding y - intercept we simply plugin x = 0 in given equation.
As we know at x = 0 , value of function is -5.
So y - intercept is -5.
Answer:
Horizontal asymptote of the graph of the function f(x) = (8x^3+2)/(2x^3+x) is at y=4
Step-by-step explanation:
I attached the graph of the function.
Graphically, it can be seen that the horizontal asymptote of the graph of the function is at y=4. There is also a <em>vertical </em>asymptote at x=0
When denominator's degree (3) is the same as the nominator's degree (3) then the horizontal asymptote is at (numerator's leading coefficient (8) divided by denominator's lading coefficient (2))
