Question 564909
Hey! Im currently in e2020 and im a visual learner NOT a person who learns just by hearing, but im in need of someone to help me on this math problem:
finding the vertex of y=x^2+3x-4
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Since this is a 2nd degree equation you should know that its graph is a parabola.
The standard form of this equation: y=A(x-h)^2+k, (h,k) being the (x,y) coordinates of the vertex. A is a multiplier which affects the slope or steepness of the curve.
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To rewrite equation in standard form, complete the square:
y=(x^2+3x+9/4)-4-9/4
y=(x+3/2)^2-45/4
This is an equation of a parabola with vertex at (-3/2,-45/4).
Since the lead coefficient>0, parabola opens upwards
see graph below as a visual check:
{{{ graph( 300, 300, -12, 12, -12, 12, (x+3/2)^2-45/4) }}}