in short, you simply pick a few random "x" values, to get the "y", and that's your point, for example say x = 2, then y = -(2)² - 4 => y = -8, that gives us the point of (2, -8), and so on.
we can start off by finding the vertex, the U-turn of the graph, and then just pick a point to its left side and a point to its right side, and we can get the vertex of that by
![\bf y=-4x^2-4\implies y=-4x^2+0x-4 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-4}x^2\stackrel{\stackrel{b}{\downarrow }}{+0}x\stackrel{\stackrel{c}{\downarrow }}{-4} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{0}{2(-4)}~,~-4-\cfrac{0}{4(-4)} \right)\implies (0~,~-4-0)\implies (0,-4)](https://tex.z-dn.net/?f=%5Cbf%20y%3D-4x%5E2-4%5Cimplies%20y%3D-4x%5E2%2B0x-4%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ctextit%7Bvertex%20of%20a%20vertical%20parabola%2C%20using%20coefficients%7D%20%5C%5C%5C%5C%20y%3D%5Cstackrel%7B%5Cstackrel%7Ba%7D%7B%5Cdownarrow%20%7D%7D%7B-4%7Dx%5E2%5Cstackrel%7B%5Cstackrel%7Bb%7D%7B%5Cdownarrow%20%7D%7D%7B%2B0%7Dx%5Cstackrel%7B%5Cstackrel%7Bc%7D%7B%5Cdownarrow%20%7D%7D%7B-4%7D%20%5Cqquad%20%5Cqquad%20%5Cleft%28-%5Ccfrac%7B%20b%7D%7B2%20a%7D~~~~%20%2C~~~~%20c-%5Ccfrac%7B%20b%5E2%7D%7B4%20a%7D%5Cright%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28-%5Ccfrac%7B0%7D%7B2%28-4%29%7D~%2C~-4-%5Ccfrac%7B0%7D%7B4%28-4%29%7D%20%5Cright%29%5Cimplies%20%280~%2C~-4-0%29%5Cimplies%20%280%2C-4%29)
and since it's a vertical parabola, the axis of symmetry comes from the x-coordinate of the vertex, namely x = 0, check the picture below.
Answer:
Step-by-step explanation:
Your question is unclear. What are "major intercepts"?
x²/81 + y²/64 = 1
The major axis is horizontal.
0²/81 + y²/64 = 1
y-intercepts = ±8
x²/81 + 0/64 = 1
x-intercepts = ±9
Direct variation is of the form y = kx where k is a constant
So its A
Answer:
uhm, i need this answer also
Step-by-step explanation:
did you ever get the answer ? if so could you be kind enough to give it to me pleasee
Answer:
Step-by-step explanation:
From the upper isosceles triangle,
The answer is: x=±√5y [Note: If and only if at the lower triangle is NOT isosceles]
Exterior angle to the isosceles triangle = 2(2x^2 – 2x) [because the exterior angle of a triangle is equal to the sum of two opposite interior angles]
Therefore, the exterior angle = 4x^2 – 4x
Also,
The exterior angle is equal to the sum of the two interior base angles of the lower triangle
Thus, 4x^2 – 4x=(3x^2+4x) + (5y – 8x)
By collecting like terms,
4x^2 – 3x^2 – 4x – 4x + 8x – 5y = 0
x^2 – 5y = 0
x^2 = 5y
By taking the square root of both sides
x=±√5y
Learn more about triangles at brainly.com/brainly.com/question/29069807
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