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Sagot :
Answer:
[tex]f(x) = \dfrac{1}{8}(x + 4)^2(x + 1)(x - 3)[/tex]
Step-by-step explanation:
From inspection of the graph, we can see that the curve intercepts the x-axis at (-4, 0), (-1, 0) and (3, 0)
Therefore,
x = -4 ⇒ x + 4 = 0
x = -1 ⇒ x + 1 = 0
x = 3 ⇒ x - 3 = 0
Because (-4, 0) touches the x-axis, then (x + 4)² will be a factor
So (x + 4)², (x + 1) and (x - 3) are all factors of the polynomial
[tex]\implies f(x) = \dfrac{1}{n}(x + 4)^2(x + 1)(x - 3)[/tex]
If we multiply the constants, this will give us the y-intercept:
⇒ 4² x 1 x -3 = -48
From inspection of the graph, the y-intercept is -6
So to get from -48 to -6 we need to multiply -48 by 1/8
Therefore, n = 1/8
[tex]\implies f(x) = \dfrac{1}{8}(x + 4)^2(x + 1)(x - 3)[/tex]
Step-by-step explanation:
the factored representation has at least one major advantage : the zero-solutions are directly visible.
as whenever one factor turns zero, the whole function turns zero.
the zero solutions are
(-4, 0)
(-1, 0)
(3, 0)
based on this we get
(x + 4)(x + 1)(x - 3)
for x = -4 the first factor is 0.
for x = -1 the second factor is 0.
for x = 3 the third factor is zero.
the y-intercept gives us a hint about 1/#, as this is the y value when x = 0.
this is then
1/# × (0+4)(0+1)(0-3) = 1/# × -12
the point (0, -6) tells us this must be equal to -6.
1/# × -12 = -6
1/# = -6/-12 = 1/2
# = 2
so, we get
f(x) = 1/2 × (x + 4)(x + 1)(x - 3)
but that is still not the whole result.
why ?
because the first zero solution (-4, 0) is actually the merger of 2 zero solutions, as the vertex of a loop is only touching the x-axis.
if we shift the whole graph a little bit down, we see that this x-intercept would turn into 2 intercepts.
so, this function has actually 4 zero solutions.
that tells us that the polynomial is of 4th degree (there must be a "x⁴" somewhere in the expression as highest exponent of x).
but multiplying only 3 factors based on x will give us only x³ as highest x exponent.
the solution : the first factor has to represent 2 zero solutions as the graph indicates.
how ?
by squaring that factor.
so, our function looks like
f(x) = 1/2 × (x + 4)²(x + 1)(x - 3)
now we get "x⁴" in the expression.
but wait, now for x = 0 the 1/# is not right anymore.
we have now
1/# × (0+4)²(0+1)(0-3) = 1/# × 16×1×-3 = 1/# × -48
so,
1/# × -48 = -6
1/# = -6/-48 = 1/8
# = 8
so, finally, the fully correct function is
f(x) = 1/8 × (x + 4)²(x + 1)(x - 3)
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