Question 1118926
<br>
{{{y=(1/2)x^2+x-4}}}<br>
Factor out the leading coefficient and complete the square in x to find the vertex and axis of symmetry.<br>
{{{y = (1/2)(x^2+2x-8)}}}
{{{y = (1/2)(x^2+2x+1-9)}}}
{{{y = (1/2)((x^2+2x+1)-9)}}}
{{{y = (1/2)((x+1)^2)-9/2}}}<br>
The vertex is (-1,-9/2); the axis of symmetry is x=-1.<br>
The y-intercept is where x=0; setting x=0 in the original form of the equation we see the y-intercept is (0,-4).<br>
Then since the axis of symmetry is x = -1, the mirror of the y-intercept is (-2,-4).<br>
To find the x-intercepts (the zeros of the function), the easiest way is to look at the original equation after the leading coefficient has been factored out:<br>
{{{y = (1/2)(x^2+2x-8)}}}
{{{y = (1/2)(x+4)(x-2)}}}<br>
The zeros are -4 and +2; the x-intercepts are (-4,0) and (2,0).<br>
{{{graph(400,400,-10,10,-10,10,(1/2)x^2+x-4)}}}