Math

• Notes
• HW 2.6
• HW 2.7
• HW 3.1
• HW 3.2
• HW 3.3
• HW 3.4
• HW 3.5
• HW 4.1
• HW 4.2
• HW 4.3
• HW 4.4
• HW 4.5
• Exam November 9th
• Exam November 9th Corrections
• Test Review 03-08-2016

• Chem 7.31 - 7.45

HW 3.1 Q4

Find the vertex.

$y = ax² + bx + c$

$f(x) = -x² + 8x + 2$

$x = \frac{-b}{2a}$

$x = \frac{-8}{2(-1)}$

$x = \frac{-8}{-2}$

$x = 4$

$y = -4² + 8(4) + 2$
$y = -16 + 32 + 2$
$y = 18$

$Vertex = (4,18)$

HW 3.1 Q5

$f(x) = 2x² - 16x - 1$

$x = \frac{-(-16)}{2(2)} = \frac{16)}{4} = 4$

$y = 2(4)² - 16(4) - 1$

$= 32 - 64 - 1$
$= -33$

Because $a > 0$, the parabola opens upward, and therefore, it has a minimum vertex, whose value is $-33$ and occurs at $x=4$.

HW 3.1 Q6

$f(x) = -3x² + 6x - 6$

$x = \frac{-6}{2(-3)}$

$x = \frac{-6}{-6} = 1$

$y = -3(1)² + 6(1) - 6$

$= -3 + 6 - 6 = -3$

Because $a < 0$, the parabola opens downward, and therefore has a maximum vertex.

The maximum value is $-3$ on the $y$-axis and occurs at $x = 1$.

The domain ($x$-axis) is negative infinity to positive infinity $(-\infty,\infty)$. The range ($y$-axis) is $(-\infty,-3]$.

HW 3.1 Q7

$f(x) = 4x^2 - 4x$

$x = \frac{-(-4)}{2(4)} = \frac{4}{8} = \frac{1}{2}$

$y = 4(\frac{1}{2})^2 - 4(\frac{1}{2})$

$= 4(\frac{1}{4}) - \frac{4}{2} = 4/4 - 4/2$

$= 1 - \frac{2}{1} = 1 - 2$
$= -1$

Because $a > 0$, the parabola opens upwards. Function has a minimum. Minimum value is $-1$, and it occurs at $x=1/2$ or $x=0.5$. The range (y-axis) is $[-1,∞)$.

HW 3.1 Q8

$f(x) = a(x)^2 + b(x) + c$

$f(x) = -0.02x² + 1.2x + 6.3$

$x = \frac{-b}{2a}$

$x = \frac{-1.2}{2(-0.02)}$

$= \frac{-1.2}{-0.04} = 30$

$y = -0.02(30)^2 + 1.2(30) + 6.3$
$= -0.02(900) + 1.2(30) + 6.3$
$= -18 + 36 + 6.3$
$= 24.3$

$x = 30, y = 24.3, a = -0.02, b = 1.2, c = 6.3$

Maximum Horizontal Distance

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-1.2+\sqrt{{1.2}^{2}-4(-0.02)(6.3)}}{2\times -0.02}$

$=\frac{-1.2+\sqrt{1.44+ 0.504}}{-0.04}$

$=\frac{-1.2+\sqrt{1.944}}{-0.04}$

$= \frac{-1.2 + 1.39427400463467}{0.04}$

$= - \frac{0.1942740046346701}{0.04}$

$= -4.856850115866752 = -4.9$

$x \lt 0$

$x=\frac{-1.2-\sqrt{{1.2}^{2}-4(-0.02)(6.3)}}{2(-0.02)}$

$= \frac{-1.2 - 1.39427400463467}{-0.04}$

$= \frac{-2.59427400463467}{-0.04}$

$= 64.85685011586675 = 64.9$

From what height was the shot released?

$x = 0$

$f(x) = a(x)^2 + b(x) + c$

$f(x) = a(0)^2 + b(0) + 6.3$

$= 0 + 0 + 6.3 = 6.3$

HW 3.1 Q9

$P = xy$
$x + y = 22$
$y = 22 - x$

$f(x) = ax^2 + bx + c$

$P = x(22-x)$
$P(x) = x(22-x)$
$= 22x - x^2$

$a = -1, b = 22, c = 0$

$x = -\frac{b}{2a}$

$x = -\frac{22}{2(-1)} = -\frac{22}{-2} = 11$

$y = 22 -11 = 11$

Maximum Product
$x \times y$
$11 \times 11 = 16$

HW 3.1 Q10

$P = xy$

$x - y = 16$
$-y = 16 - x$
$y = x - 16$

$f(x) = ax^2 + bx + c$

$P = x(x-16)$
$P(x) = x(x-16)$
$= x^2 - 16x$

$a = 1, b = 16, c = 0$

$x = -\frac{b}{2a}$

$x = -\frac{16}{2(1)} = -\frac{16}{2} = 8$

$y = 8 -16 = -8$

Maximum Product
$x \times y$
$8 \times -8 = -64$

HW 3.1 Q11

$A = xy$

Perimeter $= 2x + 2y = 440$

$2x = 440 - 2y$
$x = 220 - y$
$y = 220 - x$

$A = x(220-x)$
$A(x) = 220x-x^2$

$a = -1, b = 220, c = 0$

$x = -\frac{220}{2(-1)} = 110$

$y = 220 - 110 = 110$

HW 3.1 Q12

$Area = length \times width$
$A = xy$

$3x + 2y = 840$

$2y = 840 - 3x$

$y = 420 -\frac{3x}{2}$

$A(x) = \frac{-3}{2}x^2+420x$
$= -1.5x^2 + 420 + 0$

$a = -1.5, b = 420, c = 0$

$x = - \frac{b}{2a}$

$x = - \frac{420}{2(-1.5)}$

$= - \frac{420}{-3}$

$= 140$

$y = \frac{840-3x}{2}$

$y = \frac{840-3(140)}{2}$

$= \frac{840-420}{2}$

$= \frac{420}{2}$

$= 210$

Total Area = $A = 140 \times 210 = 29400$

HW 3.1 Q13

Cross section $= 22-2x$

$Area = A = length \times height = x \times y$

$A(x) = (x)(22-2x)$

$= -2x^2 + 22x$

$Vertex = (-\frac{b}{2a}, f(-\frac{b}{2a}))$

$a = -2, b = 22, c = 0$

$x = -\frac{22}{2(-2)}$

$= -\frac{22}{-4}$

$x = 5.5$

Depth that will maximize the cross-sectional area $= 5.5$ inches.

$A(x) = y = (x)(22-2x)$

$= (5.5)(22-2(5.5))$

$= (5.5)(22-11)$

$= 5.5 \times 11$

$y = 60.5$

HW 3.1 Q14

$f(x) = -0.1x^2 + 1.2x + 6$

$x = \frac{-b}{2a}$

$x = \frac{-1.2}{2(-0.1)}$

$= \frac{-1.2}{-0.2}$

$x = 6$

$y = -0.1(6)^2 + 1.2(6) + 6$
$= -0.1(36) + 7.2 + 6$
$= -3.6 + 7.2 + 6$

$y = 9.6$

Maximum Horizontal Distance

$f(x) = -0.1x^2 + 1.2x + 6$

When the ball hits the ground, $height = 0$

$0 = -0.1x^2 + 1.2x + 6$

$a = -0.1, b = 1.2, c = 6$

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\frac{-1.2+\sqrt{{1.2}^{2}-4(-0.1)(6)}}{2\times -0.1}$

$=\frac{-1.2+\sqrt{1.44+2.4}}{-0.2}$

$=\frac{-1.2+\sqrt{3.84}}{-0.2}$

$= \frac{-1.2+1.959591794226542}{-0.2}$

$=\frac{0.7595917942265425}{-0.2}$

$x = -3.797958971132712 = -3.8$

$x \lt 0$ …(Reject)

$x=\frac{-1.2-\sqrt{{1.2}^{2}-4(-0.1)(6)}}{2\times -0.1}$

$= \frac{-1.2-1.959591794226542}{-0.2}$

$= \frac{-3.159591794226542}{-0.2}$

$x = 15.79795897113271 = 15.8$

HW 3.1 Q15

$f(x) = ax^2 + bx + c$
$x = 2; y = 4$
$4 = a(2)^2 + b(2) + 16$
$4 = 4a + 4b + 16$

$\frac{-b}{2a} = 4$

$-b = 4(2a)$
$-b = 8a$
$b = -8a$

$4 = a(2)^2 + b(2) + 16$
$4 = 4a + (-8a)(2) + 16$
$4 = 4a - 16a + 16$
$4 = -12a + 16$
$12a = 16 - 4$
$12a = 12$
$a = 1$

$4 = 4(1) + b(2) + 16$
$4 = 4 + 2b + 16$
$4 = 2b + 20$
$2b = 4 - 20$
$2b = -16$
$b = -8$

$a = 1, b = -8, c = 16$

BACK TO TOP