Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Answer:
The answer is 5G
Step-by-step explanation:
Answer:
yes I believe so. best of luck
Answer:
(-4, -8)
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
- Terms/Coefficients
- Solving systems of equations using substitution/elimination
Step-by-step explanation:
<u>Step 1: Define Systems</u>
x - 2y = 12
5x + 3y = -44
<u>Step 2: Rewrite Systems</u>
x - 2y = 12
- [Multiplication Property of Equality] Multiply everything by -5: -5x + 10y = -60
<u>Step 3: Redefine Systems</u>
-5x + 10y = -60
5x + 3y = -44
<u>Step 4: Solve for </u><em><u>y</u></em>
<em>Elimination</em>
- Combine 2 equations: 13y = -104
- [Division Property of Equality] Divide 13 on both sides: y = -8
<u>Step 5: Solve for </u><em><u>x</u></em>
- Define original equation: x - 2y = 12
- Substitute in <em>y</em>: x - 2(-8) = 12
- Multiply: x + 16 = 12
- [Subtraction Property of Equality] Subtract 16 on both sides: x = -4
Are you supposed to find B?