How do I graph a parabola with a vertex at (3,6) with a y-intercept at 2
1 answer:
To graph
graph the point (3,6)
yintercept is on the y axis so go up 2 units from the origin and plot the point (0,2)
now the parabola
we know it has to go through points (3,6) and (0,2)
therefor we now that it opens down
we also knwo that the vertex's y value is the axis of symmetry
so we also put a point at (6,2)
draw a parabola opening down through the points (0,2) (3,6) (6,2), with (3,6) at the top/peak
see attachment
graph the point (3,6)
yintercept is on the y axis so go up 2 units from the origin and plot the point (0,2)
now the parabola
we know it has to go through points (3,6) and (0,2)
therefor we now that it opens down
we also knwo that the vertex's y value is the axis of symmetry
so we also put a point at (6,2)
draw a parabola opening down through the points (0,2) (3,6) (6,2), with (3,6) at the top/peak
see attachment
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Answer:
No Options
∠QOB=45º
Step-by-step explanation:
we have that: point O=Incenter, sides S, Q, R they form right angles with the point O, angle OBC = 15º and angle OCR=30º, find angle QOB
we know that the sum of all the internal angles of a triangle is 180º, so
α =180-30-15-90=45º and β = 180-90-α → β = 180-90-45 =45º,
Finally β=∠QOB=45º
