Question 394985
   <pre><font size = 3 color = "indigo"><b>
Hi
Solving for x
{{{7x+x(x-3)=0}}}
   x^2 + 4x = 0
  x(x+4) = 0
  x = 0
 (x+4) = 0
  x = -4
x-intercepts:  When (f(x) = 0
{{{0=7x+x(x-3)}}}   |found x values above
 Pt(0,0) and Pt(-4,0) are the x-intercepts
{{{drawing(300,300, -6, 6, -6, 6, grid(1),
circle(0, 0,0.3),
circle(-4, 0,0.3),
graph( 300, 300, -6, 6, -6, 6,   x^2 + 4x))}}}