Graph the quadratic y=x2+2x+4 . What is the axis of symmetry?
To graph the quadratic function y = x^2 + 2x + 4, we can start by finding the vertex, which will give us the axis of symmetry. The vertex of a quadratic function in the form y = ax^2 + bx + c can be calculated using the formula:
x = -b / (2a)
In this case, a = 1 and b = 2, so substituting these values into the formula, we get:
x = -2 / (2 * 1)
x = -2 / 2
x = -1
Now, to find the y-coordinate of the vertex, we substitute this value of x back into the quadratic equation:
y = (-1)^2 + 2(-1) + 4
y = 1 - 2 + 4
y = 3
So, the vertex of the quadratic function y = x^2 + 2x + 4 is (-1, 3). The axis of symmetry is a vertical line passing through this vertex.
Now, let's plot the graph:
To start, we will plot the vertex (-1, 3).
Next, we can choose some x-values and calculate the corresponding y-values to get additional points on the graph. Let's choose x = -2, -1, 0, 1, and 2 as our x-values.
For the x-value -2:
y = (-2)^2 + 2(-2) + 4
y = 4 - 4 + 4
y = 4
So, we have the point (-2, 4).
For the x-value -1 (already plotted the vertex for this):
y = (-1)^2 + 2(-1) + 4
y = 1 - 2 + 4
y = 3
So, we already plotted the point (-1, 3).
For the x-value 0:
y = (0)^2 + 2(0) + 4
y = 0 + 0 + 4
y = 4
So, we have the point (0, 4).
For the x-value 1:
y = (1)^2 + 2(1) + 4
y = 1 + 2 + 4
y = 7
So, we have the point (1, 7).
For the x-value 2:
y = (2)^2 + 2(2) + 4
y = 4 + 4 + 4
y = 12
So, we have the point (2, 12).
Now, we can plot all these points on the graph and draw a smooth curve passing through them.
The graph of the quadratic function y = x^2 + 2x + 4 is a U-shaped curve opening upwards, and its axis of symmetry is the vertical line passing through the vertex (-1, 3).
I apologize, as an AI, I am unable to draw the graph for you.