Answer:
y=a(x-p)(x-q)
y=a(x+2+√2)(x+2-√2)
passing through point (-1,1)
substitute
1=a(-1+2+√2)(-1+2-√2)
1=a(1+√2)(1-√2)
1=a(1-2)
1=a(-1)
a=1/(-1)
a=-1
y=-(x+[2+√2])(x+[2-√2])
y=-(x2+4x+2)
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Answer:
Step-by-step explanation:
Given is a triangle RST and another triangle R'S'T' tranformed from RST
Vertices of RST are (0, 0), (negative 2, 3), (negative 3, 1).
Vertices of R'S'T' are (2, 0), (0, negative 3), (negative 1, negative 1).
Comparing the corresponding vertices we find that x coordinate increased by 2 while y coordinate got the different sign.
This indicates that there is both reflection and transformation horizontally to the right by 2 units
So first shifted right by 2 units so that vertices became
(2,0) (0,3) (-1,1)
Now reflected on the line y=0 i.e. x axis
New vertices are
(2,0) (0,-3) (-1,-1)
The <em>quadratic</em> equation 3 · x² + 7 · x - 2 = 0 has a <em>positive</em> discriminant. Thus, the expression has two <em>distinct real</em> roots (<em>real</em> and <em>irrational</em> roots).
<h3>How to determine the characteristics of the roots of a quadratic equation by discriminant</h3>
Herein we have a <em>quadratic</em> equation of the form a · x² + b · x + c = 0, whose discriminant is:
d = b² - 4 · a · c (1)
There are three possibilities:
- d < 0 - <em>conjugated complex</em> roots.
- d = 0 - <em>equal real</em> roots (real and rational root).
- d > 0 - <em>different real</em> roots (real and irrational root).
If we know that a = 3, b = 7 and c = - 2, then the discriminant is:
d = 7² - 4 · (3) · (- 2)
d = 49 + 24
d = 73
The <em>quadratic</em> equation 3 · x² + 7 · x - 2 = 0 has a <em>positive</em> discriminant. Thus, the expression has two <em>distinct real</em> roots (<em>real</em> and <em>irrational</em> roots).
To learn more on quadratic equations: brainly.com/question/2263981
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