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If the roots of the equation f(x)=0 are -4,-1, 2 and 5, then binomials (x+4), (x+1), (x-2) and (x-5) are factors of needed polynomial.
Thus, the polynomial will have form
where a is real coefficient (positive or negative). This polynomial has degree 4 as needed.
The form of the graph depends on the sign of coefficient a. Attached diagrams show two different cases of possible forms of graphs (first one for positive coefficient a, second one for negative coefficient a).
First of all, remember what the equation of a line is:
Where:
m=slope
b=y-intercept.
Given this information, we can start by plugging in what we know, we are given the slope which is 2 so we can fill that in, remember m=slope
Because we were given the point (14,3)
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We need to find the value of b. The 2 is already set and x and y are variables. We want the equation for the line that specifically passes through the given point (14,3).
<span>Plug in for x the number 14 and for y the number 3:</span>(14,3).
y=mx+b or 3=2 × 14+b,
solving for b: b=3-(2)(14). b=-25.<span>Plug that in for b:
</span><span>y=2x-25
<span>Final answer:
</span><span>y=2x-25</span></span><span><span>
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