Hi, please help!
1 answer:
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Answer:
x^4 -53x^2 +108x +160
Step-by-step explanation:
If <em>a</em> is a zero, then (<em>x-a</em>) is a factor. For the given zeros, the factors are ...
p(x) = (x +8)(x +1)(x -4)(x -5)
Multiplying these out gives the polynomial in standard form.
= (x^2 +9x +8)(x^2 -9x +20)
We note that these factors have a sum and difference with the same pair of values, x^2 and 9x. We can use the special form for the product of these to simplify our working out.
= (x^2 +9x)(x^2 -9x) +20(x^2 +9x) +8(x^2 -9x) +8(20)
= x^4 -81x^2 +20x^2 +180x +8x^2 -72x +160
p(x) = x^4 -53x^2 +108x +160
_____
The graph shows this polynomial has the required zeros.
Answer:
vertex = (- 1, 1 )
Step-by-step explanation:
Given a parabola in standard form
y = ax² + bx + c ( a ≠ 0 )
Then the x- coordinate of the vertex is
x = -
y = x² + 2x + 2 ← is in standard form
with a = 1 and b = 2 , then
x = - = - 1
substitute x = - 1 into the equation for y- coordinate of vertex
y = (- 1)² + 2(- 1) + 2 = 1 - 2 + 2 = 1
vertex = (- 1, 1 )