Solved on Nov 03, 2023

Écrivez une fonction polynomiale satisfaisant les conditions: a) xx dans les quadrants III-IV, 1 point d'inflexion, yy-int: 2. b) xx dans les quadrants III-I, 3 abscisses à l'origine. c) croissante, degré 1, yy-int: -3. d) 2 points d'inflexion, yy-int: 5. e) y2y \leq 2, yy-int: 2.

STEP 1

Assumptions1. We are asked to write the equation of a polynomial function that satisfies each set of characteristics. . We are assuming that the polynomial function is in the form f(x)=axn+bxn1+...+kf(x) = ax^n + bx^{n-1} + ... + k where a,b,...,ka, b, ..., k are constants, xx is the variable, and nn is the degree of the polynomial.
3. The characteristics given in each part of the question provide information about the shape, degree, and certain points of the polynomial function.

STEP 2

For part a), we are given that the function extends from quadrant III to quadrant IV, has a turning point, and the y-intercept is2. This suggests a cubic function of the form f(x)=ax+bx2+cx+df(x) = ax^ + bx^2 + cx + d where a<0a<0 (since the function extends from quadrant III to quadrant IV), and d=2d=2 (since the y-intercept is2).
f(x)=ax+bx2+cx+2f(x) = ax^ + bx^2 + cx +2

STEP 3

For part b), we are given that the function extends from quadrant III to quadrant I, and has three x-intercepts. This suggests a cubic function of the form f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d where a>0a>0 (since the function extends from quadrant III to quadrant I), and the roots of the function are the x-intercepts. Let's say the x-intercepts are r,s,tr, s, t. Then the function can be written asf(x)=a(xr)(xs)(xt)f(x) = a(x-r)(x-s)(x-t)

STEP 4

For part c), we are given that the function is increasing, of degree1, and the y-intercept is -3. This suggests a linear function of the form f(x)=ax+bf(x) = ax + b where a>0a>0 (since the function is increasing), and b=3b=-3 (since the y-intercept is -3).
f(x)=ax3f(x) = ax -3

STEP 5

For part d), we are given that the function has two turning points, and the y-intercept is5. This suggests a quartic function of the form f(x)=ax4+bx3+cx2+dx+ef(x) = ax^4 + bx^3 + cx^2 + dx + e where e=5e=5 (since the y-intercept is5).
f(x)=ax4+bx3+cx2+dx+5f(x) = ax^4 + bx^3 + cx^2 + dx +5

STEP 6

For part e), we are given that the range of the function is y2y \leq2, and the y-intercept is2. This suggests a constant function of the form f(x)=kf(x) = k where k=2k=2 (since the y-intercept is2 and the range is y2y \leq2).
f(x)=2f(x) =2

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