Question

a) Determine o valor de K para que a função definida por f(x) = -4x² + (K + 1)x + 2 admita o valor máximo para x = 2. b) Determine o valor de P na função dada por f(x) = 3x² -2x + P para que o valor mínimo seja 5/3. c) Calcule o valor mínimo da função dada por: f(x) = 2x² -3x -2.

Answer 1

a) $f(x)=-4x^2+(k+1)x+2$

$x_V=\dfrac{-b}{2a}$

$\dfrac{-(k+1)}{2\cdot(-4)}=2$

$\dfrac{-k-1}{-8}=2$

$\dfrac{k+1}{8}=2$

$k+1=8\cdot2$

$k+1=16$

$k=16-1$

$k=15$

b) $f(x)=3x^2-2x+P$

$y_V=\dfrac{-\Delta}{4a}$

$\Delta=(-2)^2-4\cdot3\cdot P$

$\Delta=4-12P$

$\dfrac{-(4-12P)}{4\cdot3}=\dfrac{5}{3}$

$\dfrac{12P-4}{12}=\dfrac{5}{3}$

$3\cdot(12P-4)=5\cdot12$

$36P-12=60$

$36P=60+12$

$36P=72$

$P=\dfrac{72}{36}$

$P=2$

c) $f(x)=2x^2-3x-2$

$y_V=\dfrac{-\Delta}{4a}$

$\Delta=(-3)^2-4\cdot2\cdot(-2)$

$\Delta=9+16$

$\Delta=25$

$y_V=\dfrac{-25}{4\cdot2}$

$y_V=\dfrac{-25}{8}$

Answer 2

A) f(x) = -4x² + (k + 1)x + 2

$x_{v} = \frac{ - b}{2a} \\ 2 = \frac{ - (k + 1)}{2 \times - 4} \\ 2 = \frac{ - k - 1}{ - 8} \\ - k - 1 = 2 \times - 8 \\ - k - 1 = - 16 \\ k + 1 = 16 \\ k = 16 - 1 \\ k = 15$

b) f(x) = 3x² - 2x + P

$y_{v} = \frac{ - \Delta}{4a} \\ \\ \Delta = {2}^{2} - 4 \times 3 \times p \\ \Delta = 4 - 12p \\ \\ \frac{5}{3} = - \frac{4 - 12p}{4 \times 3 } \\ \frac{5}{3} = - \frac{4 - 12p}{12} \\ 60 = - 3( 4 - 12p) \\ 20 = - 4 + 12p \\ p = \frac{24}{12} \\ p = 2$

c) f(x) = 2x² - 3x - 2

$\Delta = {3}^{2} - 4 \times 2 \times -2 \\ \Delta = 9 + 16 \\ \Delta = 25 \\ \\ y = \frac{ - 25}{4 \times 2} \\ y = \frac{ - 25}{8}$

Espero Ter Ajudado !!