Asked by Lisa
Can you help me with the following questions, I don't understand. thanks!
Write the equation of a polynomial function with the given characteristics. Use a leading coefficient of 1 or -1 and make the degree of the function as small as possible.
#1. Crosses the x-axis at -4, 0, and 1; lies above the x-axis between -4 and 0; lies below the x-axis between 0 and 1.
#2.Touches the x-axis at 0 and crosses the x-axis at 4; lies below the x-axis between 0 and 4
Write the equation of a polynomial function with the given characteristics. Use a leading coefficient of 1 or -1 and make the degree of the function as small as possible.
#1. Crosses the x-axis at -4, 0, and 1; lies above the x-axis between -4 and 0; lies below the x-axis between 0 and 1.
#2.Touches the x-axis at 0 and crosses the x-axis at 4; lies below the x-axis between 0 and 4
Answers
Answered by
Reiny
for #1
x-intercepts of a function are solutions to the corresponding equation
since there are solutions of -4,0,and 1
the function must have been f(x) = (x+4)x-0)(x-1)
or
f(x) = x(x-1)(x+4) . You can expand it or leave it as is
for #2
if a graph "touches" the x-axis then there must have been a double root, such as (x-2)(x-2)
so this one clearly must have been
f(x) = (x-0)(x-0)(x-4)
= x^2(x-4)
In both cases we need a cubic for 3 roots. If the x^3 is positive, then the cubic "rises" into the first quadrant and "drops" into the third quadrant.
Since this was the case for both functions there would be a +1 understood in front of the factors,
had it been reversed, a -1 in front of the factors would have achieved this
x-intercepts of a function are solutions to the corresponding equation
since there are solutions of -4,0,and 1
the function must have been f(x) = (x+4)x-0)(x-1)
or
f(x) = x(x-1)(x+4) . You can expand it or leave it as is
for #2
if a graph "touches" the x-axis then there must have been a double root, such as (x-2)(x-2)
so this one clearly must have been
f(x) = (x-0)(x-0)(x-4)
= x^2(x-4)
In both cases we need a cubic for 3 roots. If the x^3 is positive, then the cubic "rises" into the first quadrant and "drops" into the third quadrant.
Since this was the case for both functions there would be a +1 understood in front of the factors,
had it been reversed, a -1 in front of the factors would have achieved this
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