Quadratic form to vertex form with a vertex of (-4,-1) and a y-intercept of 7
1 answer:
The quadratic function in vertex form is:
y = a(x - h)^2 + k
Where:
vertex = (h, k)
Axis of symmetry: x = h
The value of “a” determines whether the graph opens up or down, and makes the parent function wider or narrower.
The value of “h” determines how far left or right the parent function is translated.
The value of “k” determines how far up or down the parent function is translated.
Now that we have these definitions, we can substitute the given values into the vertex form to solve for “a”:
Use vertex = (-4, -1) and y-intercept, (0, 7):
7 = a(0+ 4)^2 - 1
7 = a(4)^2 - 1
7 = a(16) - 1
Add 1 to both sides:
7 + 1 = a(16) - 1 + 1
8 = 16a
Divide both sides by 16 to solve for “a”:
8/16 = 16a/16
1/2 = a
Since a = 1/2 (which is positive, implying that the parabola opens upward), and the vertex occurs at point (-4, -1) as the minimum point:
The quadratic equation in vertex form is:
y = 1/2(x + 4)^2 - 1
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y = a(x - h)^2 + k
Where:
vertex = (h, k)
Axis of symmetry: x = h
The value of “a” determines whether the graph opens up or down, and makes the parent function wider or narrower.
The value of “h” determines how far left or right the parent function is translated.
The value of “k” determines how far up or down the parent function is translated.
Now that we have these definitions, we can substitute the given values into the vertex form to solve for “a”:
Use vertex = (-4, -1) and y-intercept, (0, 7):
7 = a(0+ 4)^2 - 1
7 = a(4)^2 - 1
7 = a(16) - 1
Add 1 to both sides:
7 + 1 = a(16) - 1 + 1
8 = 16a
Divide both sides by 16 to solve for “a”:
8/16 = 16a/16
1/2 = a
Since a = 1/2 (which is positive, implying that the parabola opens upward), and the vertex occurs at point (-4, -1) as the minimum point:
The quadratic equation in vertex form is:
y = 1/2(x + 4)^2 - 1
Please mark my answers as the Brainliest, if you find my explanations/solution helpful :)
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Given <em>z</em> = 3 + <em>i</em>, right away we can find
(a) square
<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>
(b) modulus
|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10
(d) polar form
First find the argument:
arg(<em>z</em>) = arctan(1/3)
Then
<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))
<em>z</em> = √10 exp(<em>i</em> arctan(1/3))
or
<em>z</em> = √10 (cos(arctan(1/3)) + <em>i</em> sin(arctan(1/3))
(c) square root
Any complex number has 2 square roots. Using the polar form from part (d), we have
√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)
and
√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 2)
Then in standard rectangular form, we have
and
We can simplify this further. We know that <em>z</em> lies in the first quadrant, so
0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2
which means
0 < 1/2 arctan(1/3) < <em>π</em>/4
Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have
and since cos(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),
Now, arctan(1/3) is an angle <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 1/√10. Then
So the two square roots of <em>z</em> are
and