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Rewrite the equation in the slope-intercept form y = mx + c.

Mathematics

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*1) Given the equation of a line: x4+y3−12=0\frac{x}{4} + \frac{y}{3} - \frac{1}{2} = 0.

a) Determine the gradient of the line.

  1. Rewrite the equation in the slope-intercept form y=mx+cy = mx + c.

    y3=−x4+12\frac{y}{3} = -\frac{x}{4} + \frac{1}{2}

  2. Multiply by 3 to isolate yy.

    y=3(−x4+12)y = 3 \left(-\frac{x}{4} + \frac{1}{2}\right) y=−34x+32y = -\frac{3}{4}x + \frac{3}{2}

  3. Identify the gradient mm.

    The gradient mm is the coefficient of xx.

    m=−34m = -\frac{3}{4}

    The gradient of the line is −34\boxed{-\frac{3}{4}}.

    b) Determine the coordinates of the point where the line cuts the coordinate axes.

  4. Find the x-intercept (where y=0y=0).

    Substitute y=0y=0 into the original equation:

    x4+03−12=0\frac{x}{4} + \frac{0}{3} - \frac{1}{2} = 0 x4−12=0\frac{x}{4} - \frac{1}{2} = 0 x4=12\frac{x}{4} = \frac{1}{2} x=4×12x = 4 \times \frac{1}{2} x=2x = 2

    The x-intercept is (2,0)(2, 0).

  5. Find the y-intercept (where x=0x=0).

    Substitute x=0x=0 into the original equation:

    04+y3−12=0\frac{0}{4} + \frac{y}{3} - \frac{1}{2} = 0 y3−12=0\frac{y}{3} - \frac{1}{2} = 0 y3=12\frac{y}{3} = \frac{1}{2} y=3×12y = 3 \times \frac{1}{2} y=32y = \frac{3}{2}

    The y-intercept is (0,32)\left(0, \frac{3}{2}\right).
    The coordinates where the line cuts the coordinate axes are (2,0) and (0,32)\boxed{(2, 0) \text{ and } \left(0, \frac{3}{2}\right)}.

    c) An equation of a line passing through (a,b)(a, b) and perpendicular to the line above.

  6. Find the gradient of the given line.

    From part (a), the gradient of the given line is m=−34m = -\frac{3}{4}.

  7. Find the gradient of the perpendicular line.

    If two lines are perpendicular, the product of their gradients is −1-1.
    Let m⊥m_{\perp} be the gradient of the perpendicular line.

    m⋅m⊥=−1m \cdot m_{\perp} = -1 (−34)m⊥=−1\left(-\frac{3}{4}\right) m_{\perp} = -1 m⊥=−1−34m_{\perp} = \frac{-1}{-\frac{3}{4}} m⊥=43m_{\perp} = \frac{4}{3}

  8. Use the point-slope form y−y1=m⊥(x−x1)y - y_1 = m_{\perp}(x - x_1) with (x1,y1)=(a,b)(x_1, y_1) = (a, b).

    y−b=43(x−a)y - b = \frac{4}{3}(x - a)

  9. Rearrange into a standard form (optional, but often preferred).

    3(y−b)=4(x−a)3(y - b) = 4(x - a) 3y−3b=4x−4a3y - 3b = 4x - 4a 4x−3y−4a+3b=04x - 3y - 4a + 3b = 0

    The equation of the perpendicular line is y−b=43(x−a) or 4x−3y−4a+3b=0\boxed{y - b = \frac{4}{3}(x - a) \text{ or } 4x - 3y - 4a + 3b = 0}.

    2) The price tag on an article in a supermarket is 7,000 FCFA. During an auction sale, the price was reduced by 12%.

    a) At what price will the article be sold?

  10. Calculate the reduction amount.

    Reduction = 12%12\% of 7,000 FCFA

    Reduction=12100×7000\text{Reduction} = \frac{12}{100} \times 7000 Reduction=12×70\text{Reduction} = 12 \times 70 Reduction=840 FCFA\text{Reduction} = 840 \, \text{FCFA}

  11. Calculate the selling price.

    Selling price = Original price - Reduction

    Selling price=7000−840\text{Selling price} = 7000 - 840 Selling price=6160 FCFA\text{Selling price} = 6160 \, \text{FCFA}

    The article will be sold for 6160 FCFA\boxed{6160 \, \text{FCFA}}.

    b) Given that there is a 1% tax on the purchase price, how much will be paid? (Leave your answer to the nearest 5FCFA).

  12. Calculate the tax amount.

    The purchase price is the selling price from part (a), which is 6160 FCFA.
    Tax = 1%1\% of 6160 FCFA

    Tax=1100×6160\text{Tax} = \frac{1}{100} \times 6160 Tax=61.60 FCFA\text{Tax} = 61.60 \, \text{FCFA}

  13. Calculate the total amount paid.

    Total paid = Selling price + Tax

    Total paid=6160+61.60\text{Total paid} = 6160 + 61.60 Total paid=6221.60 FCFA\text{Total paid} = 6221.60 \, \text{FCFA}

  14. Round the total amount to the nearest 5 FCFA.

    The nearest multiple of 5 to 6221.60 is 6220.
    The amount to be paid is 6220 FCFA\boxed{6220 \, \text{FCFA}}.

    3) Given that f(x)=x3−2x2−5x+6f(x) = x^3 - 2x^2 - 5x + 6.

    a) Show that (x+2)(x+2) is a factor of ff.

  15. Use the Factor Theorem.

    According to the Factor Theorem, if (x+2)(x+2) is a factor of f(x)f(x), then f(−2)f(-2) must be equal to 0.
    Substitute x=−2x = -2 into f(x)f(x):

    f(−2)=(−2)3−2(−2)2−5(−2)+6f(-2) = (-2)^3 - 2(-2)^2 - 5(-2) + 6 f(−2)=−8−2(4)−(−10)+6f(-2) = -8 - 2(4) - (-10) + 6 f(−2)=−8−8+10+6f(-2) = -8 - 8 + 10 + 6 f(−2)=−16+16f(-2) = -16 + 16 f(−2)=0f(-2) = 0

    Since f(−2)=0f(-2) = 0, (x+2) is a factor of f(x)\boxed{(x+2) \text{ is a factor of } f(x)}.

    b) Factorise f(x)f(x) completely.

  16. Divide f(x)f(x) by (x+2)(x+2) to find the quadratic factor.

    Using synthetic division with root −2-2:

    -2 | 1   -2   -5    6
   |     -2    8   -6
   ------------------
     1   -4    3    0

    The coefficients of the quotient are 1,−4,31, -4, 3.
    So, the quadratic factor is x2−4x+3x^2 - 4x + 3.

  17. Factorise the quadratic expression x2−4x+3x^2 - 4x + 3.

    We need two numbers that multiply to 3 and add to -4. These numbers are -1 and -3.

    x2−4x+3=(x−1)(x−3)x^2 - 4x + 3 = (x - 1)(x - 3)

  18. Write f(x)f(x) as a product of its linear factors.

    f(x)=(x+2)(x−1)(x−3)f(x) = (x+2)(x-1)(x-3)

    The complete factorisation of f(x)f(x) is (x+2)(x−1)(x−3)\boxed{(x+2)(x-1)(x-3)}.

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