y= 4c+3
and assuming it is multiple choice, the graph should look similar to that. Hope that helps!
I need help with this
2 answers:
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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)
