The x-intercept is a point on the graph that is located on any point of x but must have a y-value of 0. To find the x-intercept, we must set y in the given equation equal to 0.

8x + 2y = 4

8x + 2(0) = 4

8x = 4

x = 4/8

x = 1/2

4 0

1 year ago

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

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$

7 0

3 years ago

W(1,8),X(7,8),Y(4,5), and Z(1,2) select all methods you can use to prove WYZ is congruent to WYX

marin [14]

The distance between two points (x₁,y1),(x₂,y₂) is d
$d= \sqrt{(x2-x1)^2 + (y2-y1)^2}$

For the given problem we have
W(1,8),X(7,8),Y(4,5), and Z(1,2)

The length of WY = $\sqrt{(4-1)^2 + (5-8)^2}$ = 3√2
The length of WX = $\sqrt{(7-1)^2 + (8-8)^2}$ = 6
The length of WZ = $\sqrt{(1-1)^2 + (2-8)^2}$ = 6
The length of XY = $\sqrt{(4-7)^2 + (5-8)^2}$ = 3√2
The length of ZY = $\sqrt{(1-4)^2 + (2-5)^2}$ = 3√2

∴ WX = WZ ⇒⇒⇒ proved
XY = ZY ⇒⇒⇒ proved
WY = WY ⇒⇒⇒ reflexive property

∴ ΔWYZ is congruent to ΔWYX by SSS method

5 0

3 years ago

LetFfor a fractional linear map. Letz′i=F(zi) fori= 1,2,3,4. Show that thecross ratio ofz′1, z′2, z′3, z′4is the same as the cro

Dovator [93]

Answer:

Just try OK

Step-by-step explanation:

5 0

3 years ago