Question 1187844

Given the quadratic function:

{{{f(x) = -(1/2)x^2-3x+5/2}}}

a. Determine whether the parabola opens up or opens down.

if the leading coefficient is less than zero, as in your case, 
the parabola opens {{{down}}}

b. Find the vertex of the parabola.

rewrite equation in vertex form

{{{f(x) = (-(1/2)x^2-3x)+5/2}}}

{{{f(x) = -(1/2)(x^2+6x)+5/2}}}

{{{f(x) = -(1/2)(x^2+6x+b^2)-(-1/2)b^2+5/2}}}..........{{{b=6/2=3}}}

{{{f(x) = -(1/2)(x^2+6x+3^2)+(1/2)3^2+5/2}}}

{{{f(x) = -(1/2)(x+3)^2+9/2+5/2}}}

{{{f(x) = -(1/2)(x+3)^2+7}}}=>{{{h=-3}}} and {{{k=7}}}

the vertex is at ({{{-3}}}, {{{7}}})


c. Determine the axis of the symmetry.

The {{{x}}} -coordinate of the vertex is the equation of the axis of symmetry of the parabola.

{{{x=-3}}}

d. Find the x - intercepts and the y - intercept of the parabola.

x - intercepts: set {{{f(x) =0}}}

{{{0 = -(1/2)(x+3)^2+7}}}.........solve for {{{x}}}

{{{ (1/2)(x+3)^2=7}}}

{{{ (x+3)^2=7/(1/2)}}}

{{{ (x+3)^2=14}}}

{{{ x+3=sqrt(14)}}}
{{{ x=sqrt(14) -3}}}

solutions:
{{{x =  sqrt(14)-3}}}-> approximately {{{0.7}}}
{{{x =  - sqrt(14)-3}}}-> approximately {{{-6.7}}}

y - intercepts: set {{{x =0}}}

recall that {{{f(x)=y}}}

{{{y= -(1/2)(0+3)^2+7}}}
{{{y= 5/2}}}

e. Sketch the graph of f(x).


{{{ graph( 600, 600, -10, 10, -10, 10, -(1/2)(x+3)^2+7) }}}