Answer:
the vertex of this parabola is at (5/2, 47/4)
Step-by-step explanation:
Y=x^2+5x-7 is written in 'standard form' for a quadratic. Comparing this to
y = ax^2 + 5x - 7, we see that a = 1 and b = 5.
-b
The formula for the x-coordinate of the vertex is x = --------
5 2a
which in this case has the value x = ------- = 5/2
2(1)
Now use the quadratic formula to find the y-value of the vertex:
y = (5/2)^2 + 5(5/2) - 7, or
y = 25/4 + 25/2 - 7, or
y = 11.75 or 11 3/4
Thus the vertex of this parabola is at (5/2, 47/4)
