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y = x2 + 6x + 7

a = 1, b = 6, c = 7

Since a is positive, the graph opens upward. You can find the vertex coordinates (-b/2a, f(-b/2a)) = (-3, f(-3)) = (-3, -2), draw the axis of symmetry, x = -b/2a = -3, plot the y-intercept point (0, 7), plot the point (-b/a, 7) = (-6, 7), and draw the graph that passes through these points.

Or complete the square.

y = x2 + 6x + 7

y = x2 + 6x + 9 - 9 + 7

y = (x2 + 6x + 9) - 2

y = (x + 3)2 - 2

So start with the graph of y = x2 whose vertex is at the origin. Move it 3 units to the left, and 2 units down.

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14y ago
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Q: How to Graphs Y equals x2 plus 6x plus 7?
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