Question

What is the vertex of the graph of #y=x^2+3x-4#?

Answer 1

To find the x value of the vertex we do:

x = $\frac{-b}{2a}$
if your quadratic function is in the form ax² + bx + c = y

x = $\frac{-3}{2(1)}$
x = $\frac{-3}{2}$

Now we plug this x into the equation and solve for y.

y = $(\frac{-3}{2}) ^{2} + 3( \frac{-3}{2} ) + 4$
y = $\frac{9}{4} - \frac{9}{2} + 4$
y = $\frac{9}{4} - \frac{18}{4} + 4$
y = $\frac{-9}{4} + 4$
y = $\frac{-9}{4} + \frac{16}{4}$
y = $\frac{7}{4}$

Your vertex (if my math is correct) is at $( \frac{-3}{2}, \frac{7}{4} )$