You have several approaches available.
color(blue)("Axis of symmetry - a sort of cheat way")Axis of symmetry - a sort of cheat way
Let me show you a quick way of determining the axis of symmetry.
Standardised form of y=ax^2+bx+cy=ax2+bx+c
Write as
y=a(x^2+b/ax) +c color(white)("ddd") ->color(white)("ddd") y=1(x^2+3/1x)-5y=a(x2+bax)+cddd→dddy=1(x2+31x)−5
This is the beginnings if completing the square
x_("vertex")="axis of symmetry" = (-1/2)xxb/a xvertex=axis of symmetry=(−12)×ba
color(magenta)(x_("vertex")="axis of symmetry" = (-1/2)xx3=-3/2)xvertex=axis of symmetry=(−12)×3=−32
Another way is the determine the x-intercepts (if there is any) the axis of symmetry is 1/2 way between those 2 points.
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color(blue)("Determine the x-intercepts")Determine the x-intercepts
y=x^2+3x-5y=x2+3x−5
x=(-b+-sqrt(b^2-4ac))/(2a) -> x=(-3+-sqrt(3^2-4(1)(-5)))/(2(1))x=−b±√b2−4ac2a→x=−3±√32−4(1)(−5)2(1)
x=-3/2+-sqrt(29)/2x=−32±√292
x=-4.19258.... and x=+1.19258...
x_1~~-4.19 and x_2~~1.19 to 2 decimal places
Just to check the axis of symmetry
color(magenta)((x_1+x_2)/2= -1.5=-3/2 larr" confirmed")
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color(blue)("Determine the y-intercepts")
This occures at x=0
y_("intercept")=0^2+3(0)-5 = -5
