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sveta [45]
3 years ago
5

How to change from standard for to vertex form

Mathematics
2 answers:
zalisa [80]3 years ago
8 0
If the equation is y = 3(x + 4)2<span> - 6, the value of h is -4, and k is -6. To convert a quadratic from y = ax</span>2<span> + bx + c </span>form to vertex form, y = a(x - h)2+ k, you use the process of completing the square. Let's see an example. Convert y = 2x2<span> - 4x + 5 into </span>vertex form<span>, and state the </span>vertex<span>.</span>
Arlecino [84]3 years ago
8 0
Standard form is
ax^2+bx+c=y
to change to vertex form
complete the square

HOW TO COMPLETE THE SQUARE
first, isolate the x terms
(ax^2+bx)+c=y
factor out a
a(x^2+(b/a)x)+c=y
take 1/2 of the coefient of the x term and square it
(b/a) time 1/2=b/(2a), square it, \frac{b^2}{4a^2}
now add positive and negative inside parenthasees
a(x^2+(b/a)x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2})+c=y
factor perfect square
a(((x+ \frac{b^2}{4a^2})^2-a\frac{b^2}{4a^2})+c=y
distribute
a(x+ \frac{b^2}{4a^2})^2 -a \frac{b^2}{4a^2}+c=y
 a(x+ \frac{b^2}{4a^2})^2 - \frac{b^2}{4a}+c=y
that is vertex form and how to complete the square

for ax^2+bx+c=y
vertex form is
 a(x+ \frac{b^2}{4a^2})^2 - \frac{b^2}{4a}+c=y
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Elden [556K]

Answer:

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3 years ago
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zavuch27 [327]

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Sav [38]

Answer:

FALSE

Step-by-step explanation:

<E in ∆AED ≅ <E in ∆CEB.

Both are 90°.

Side ED ≅ Side EB

Side AD ≅ Side CB.

Thus, two sides (ED and AD) and a non-included angle (<E) of ∆AED are congruent to corresponding two sides (EB and CB) and a non-included angle (<E) of ∆CEB. Therefore, by A-S-S Congruence Theorem, both triangles are congruent to each other not by SSS.

8 0
2 years ago
What is 42÷(15-2 by the power of 3<br><img src="https://tex.z-dn.net/?f=42%20%5Cdiv%2815%20-%202%20%7B%7D%5E%7B3%7D%20%29" id="T
tigry1 [53]
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4 0
3 years ago
Find all solutions in the interval (0,2pi) you should get kpi and pi/4 but how do you get there
Ket [755]

Add sin(x) to both sides

tan(x)sin(x)=sin(x)

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tan(x)=1

now you just have to figure out where tan(x)=1. you can figure this out from there by looking at the unit circle.

4 0
3 years ago
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