The roots of the entire <em>polynomic</em> expression, that is, the product of p(x) = x^2 + 8x + 12 and q(x) = x^3 + 5x^2 - 6x, are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.
<h3>How to solve a product of two polynomials </h3>
A value of <em>x</em> is said to be a root of the polynomial if and only if <em>r(x) =</em> 0. Let be <em>r(x) = p(x) · q(x)</em>, then we need to find the roots both for <em>p(x)</em> and <em>q(x)</em> by factoring each polynomial, the factoring is based on algebraic properties:
<em>r(x) =</em> (x + 6) · (x + 2) · x · (x² + 5 · x - 6)
<em>r(x) =</em> (x + 6) · (x + 2) · x · (x + 3) · (x + 2)
r(x) = x · (x + 2)² · (x + 3) · (x + 6)
By direct inspection, we conclude that the roots of the entire <em>polynomic</em> expression are <em>x₁ =</em> 0, <em>x₂ =</em> -2, <em>x₃ =</em> -3 and <em>x₄ =</em> -6.
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Answer: Choice A
y = -3(x+2)^2 + 10
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Work Shown:
y = -3x^2-12x-2 is in the form y = ax^2+bx+c with
a = -3
b = -12
c = -2
The x coordinate of the vertex is
h = -b/(2a)
h = -(-12)/(2*(-3))
h = 12/(-6)
h = -2
We'll plug this into the original equation to find the corresponding y coordinate of the vertex.
y = -3x^2-12x-2
y = -3(-2)^2-12(-2)-2
y = 10
So k = 10 is the y coordinate of the vertex.
Overall, the vertex is (h,k) = (-2,10)
Meaning that we go from this general vertex form
y = a(x-h)^2 + k
to this
y = -3(x - (-2))^2 + 10
y = -3(x+2)^2 + 10
Answer:
Step-by-step explanation:
for short lengths try your hand span.
for something a bit longer try using the length of your feet.
for something even longer try using your arm span or steps.
F(x): slope is 1.5
Y intercept is 1
G(x): slope is 2
Y intercept is 1
Statement A is correct
f(x) slope is less than g(x)
The y intercept of f(x) is the equal to the y intercept of g(x)