Home > Precalculus > Vertex Form of the Equation

How to answer these two questions ?

Answer 1

#"see explanation"#

#(a)#

#"the equation of a parabola in "color(blue)"vertex form"# is.

#color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))# where (h , k ) are the coordinates of the vertex and a is a constant.

#"the x-coordinate of the vertex is midway between the zeros"#

#rArrh=(0-4)/2=-2#

#rArrf(x)=a(x+2)^2-6#

#rArrp=k=-6#

#rArrcolor(magenta)"vertex "=(-2,-6)#

#"to find a use the point "(-4,0)" a point on the parabola"#

#"substitute into "f(x)=a(x+2)^2-6#

#0=4a-6rArra=6/4=3/2#

#(b)#

#"under a reflection in the x-axis"#

#• " a point "(x,y)to(x,-y)#

#rArr(-2,-6)to(-2,6)#

#(-4,0)" and " (0,0) " remain unchanged"#

#"under reflection the parabola will now be maximum"#

#rArra=-3/2#

#rArrf(x)=-3/2(x+2)^2+6larrcolor(red)" in vertex form"#

#rArrf(x)=-3/2x^2-6xlarrcolor(red)" in terms of x"# graph{(y-3/2x^2-6x)(y+3/2x^2+6x)=0 [-20, 20, -10, 10]}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.