Question

The area of the triangle formed by the x- and y- intercepts of the parabola y = 0.5(x-3)(x+k) is equal to 1.5 square units. Find all possible values of k.

I've already solved this question, this is just for future RSM students looking for help.

Answer 1

`k=-2, \quad k=-1, \quad k=(-3-√(17))/(2), \quad k=(-3+√(17))/(2)`

Explanation:

Given equation of the parabola:

`y = 0.5(x-3)(x+k)`

x-intercepts

The x-intercepts are the points at which the curve crosses the x-axis.

To find the x-intercepts, substitute y = 0 into the given equation and solve for x:

`\implies 0.5(x-3)(x+k)=0`

`\implies (x-3)(x+k)=0`

Apply the zero-product property:

`\implies x-3=0 \implies x=3`

`\implies x+k=0 \implies x=-k`

Therefore, the x-intercepts are:

• (3, 0) and (-k, 0)

y-intercept

The y-intercept is the point at which the curve crosses the y-axis.

To find the y-intercept, substitute x = 0 into the given equation and solve for y:

`\implies y = 0.5(0-3)(0+k)`

`\implies y = 0.5(-3)(k)`

`\implies y = -1.5k`

Therefore, the y-intercept is:

• (0, -1.5)

Area

`\boxed{\begin{minipage}{4 cm}\underline{Area of a triangle} \\\\$A=(1)/(2)bh$\\\\where:\\ \phantom{ww}$\bullet$ $b$ is the base. \\ \phantom{ww}$\bullet$ $h$ is the height. \\\end{minipage}}`

The height of the triangle is the distance between the y-intercept and the x-axis:

`\implies h=-1.5k`

The base of the triangle is the difference between the x-intercepts:

`\implies b=3+k\quad \textsf{or} \quad b=-(3+k)`

Given the area of the triangle is equal to 1.5 square units, substitute the values of b and h into the formula for area and solve for k:

For b = (3 + k):

`\begin{aligned}\implies\textsf{Area}&=(1)/(2)bh\\1.5&=(1)/(2) \cdot (3+k)\cdot(-1.5k)\\-2&=(3+k)k\\k^2+3k+2&=0\\k^2+k+2k+2&=0\\k(k+1)+2(k+1)&=0\\(k+2)(k+1)&=0\\\\ k+2&=0 \implies k=-2\\k+1&=0 \implies k=-1\end{aligned}`

For b = -(3 + k):

`\begin{aligned}\implies\textsf{Area}&=(1)/(2)bh\\1.5&=(1)/(2) \cdot -(3+k)\cdot(-1.5k)\\-2&=-(3+k)k\\2&=(3+k)k\\k^2+3k&=2\\k^2+3k+\dfra&=4.25\\\left(k+1.5\right)^2&=4.25\\k+1.5&=√(4.25)\\k+1.5&=\pm(√(17))/(2)\\k&=(-3\pm√(17))/(2)\end{aligned}`

Solution

Therefore, the possible values of k are:

`k=-2, \quad k=-1, \quad k=(-3-√(17))/(2), \quad k=(-3+√(17))/(2)`

Answer 2

{-2, - 1, (- 3 ± √17)/2}

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The vertices are

x-intercepts, at y = 0

• (x - 3)(x + k) = 0
• x = 3 and x = - k

y-intercept, at x = 0

• y = 0.5(-3)k= - 1.5k

The base of the triangle is the distance between the x-intercepts

• b = ± (k + 3), depending on value of k, and position relevant to x = 3

The height is h = - 1.5k, as it represents the distance from the x-axis

The area of the triangle

• A = bh/2
• A = 1.5

Substitute values of b and h and solve for k

• A = ± (k + 3)×(- 1.5k)/2
• 1.5 = ± 1.5k(x + 3)/2
• k( k + 3) = ± 2

Solve the quadratic equation

• k² + 3k ± 2 = 0
• D = 3² ± 2*4 = 9 ± 8

1) D = 1

• k = (-3 ± 1)/2
• k = -2, k = - 1

2) D = √17

• k = (- 3 ± √17)/2