Please help! Which choice belongs in space number 3?
2 answers:
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Considering the vertex of the parabola, the correct statement is given by:
The range of the function is all real numbers less than or equal to 9.
<h3>What is the vertex of a quadratic equation?</h3>A quadratic equation is modeled by:
The vertex is given by:
In which:
Considering the coefficient a, we have that:
- If a < 0, the vertex is a maximum point, which means that the range is all real numbers less than or equal to
.
- If a > 0, the vertex is a minimum point, which means that the range is all real numbers greater than or equal to
In this problem, we have that:
- a = -1 < 0, hence the vertex is a maximum point.
- The vertex is (-2,9).
Hence the range is described by:
The range of the function is all real numbers less than or equal to 9.
More can be learned about the vertex of a parabola at brainly.com/question/24737967
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Vertex form is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k
we have some points
(3,-2) and (6,-26)
input and solve for a and k
(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k
(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k
we have
-2=a+k
-26=4a+k
multiply first equation by -1 and add to second
2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k
-24=3a
divide both sides by 3
-8=a
-2=a+k
-2=-8+k
add 8 to both sides
6=k
the equation is
y=a(x-h)^2+k
vertex is (h,k)
axis of symmetry is x=4, therfor h=4
y=a(x-4)^2+k
we have some points
(3,-2) and (6,-26)
input and solve for a and k
(3,-2)
-2=a(3-4)^2+k
-2=a(-1)^2+k
-2=a(1)+k
-2=a+k
(6,-26)
-26=a(6-4)^2+k
-26=a(2)^2+k
-26=a(4)+k
-26=4a+k
we have
-2=a+k
-26=4a+k
multiply first equation by -1 and add to second
2=-a-k
<u>-26=4a+k +</u>
-24=3a+0k
-24=3a
divide both sides by 3
-8=a
-2=a+k
-2=-8+k
add 8 to both sides
6=k
the equation is