Anyone know the answer to this question?
1 answer:
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Answer:
The answers for your two problems, can be found below in the attached images.
Problem 1
Graph
The graph was plotted using a calculator. We can see that the graph opens up.
f(x) = x^2 + 4x +3
Vertex
The vertex is the minimum point of the equation
In this case Vertex (V) = (-2,-1)
Axis of symmetry
The x axis corresponding to the vertex component
x = -2
y intercept
Interception with the y-axis
From the first attached image, we can see that they y intercept occurs
at x = 0, y = 3
(0,3)
Problem 2
Graph.
The graph was plotted using a calculator. We can see that the graph opens up.
f(x) = 2x^2 + 3x +1
Vertex
The vertex is the minimum point of the equation
In this case Vertex (V) = (-0.75,-0.125)
Axis of symmetry
The x axis corresponding to the vertex component
x = -0.75
y intercept
Interception with the y-axis
From the second attached image, we can see that they y intercept occurs
at x = 0, y = 1
(0,1)
Hmmm the y-intercept is at -2? what does that mean? well, is where the graph "intercepts" or touches the y-axis, and when that happens, x = 0, so the point is really ( 0 , -2 ).
and we know where the vertex is at. Let's assume a vertical parabola, in which case the squared variable is the "x".
and we know where the vertex is at. Let's assume a vertical parabola, in which case the squared variable is the "x".