Given:#" "y=-4x^2+8x+13#
#color(blue)("Determine the general shape of the curve")#
The coefficient of #x^2# is negative. Consequently the graph is pf form #nn#. Thus the vertex is a maximum
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#color(blue)("Determining y-intercept")#
Reading directly off the equation the y-intercept is +13
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#color(blue)("Determining the vertex")#
Write as #-4(x^2-8/4x)+13#
#x_("vertex")=(-1/2)xx(-8/4) = +1#
By substitution: #" "y_("vertex")=-4(1)^2+8(1)+13 = 17#
Vertex#->(x,y)=(1,17)#
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#color(blue)("Determining x-intercept")#
The vertex is above the x-axis so as the general form is #nn# x-intercepts exist.
Compare to the standardised form #y=ax^2bx+c#
Where #x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=-4"; "b=8"; "c=13# giving:
#x=(-8+-sqrt(8^2-4(-4)(13)))/(2(-4))#
#x=(-8+-sqrt(272))/(-8)#
#x=1+-(sqrt(2^2xx2^2xx17))/(-8)#
#x=1+-(4sqrt(17))/-8" " ->" " 1+-sqrt(17)/2 larr" exact answer"#
You may write #4/(-8) # as #1/2# as its sign does not matter due to the #+- # preceding that part of the equation.
#x~~ 3.062" "x~~-1.062#
