Question 1161266
<pre>

y = {{{ x-4}}}
y = {{{2x^2+8x-1}}}

At the point(s) of intersection,  {{{ x-4 }}} = {{{ 2x^2+8x-1}}}

Simplifying...
{{{ 2x^2 + 7x + 3 = 0 }}}

Factoring...
{{{ (2x+1)(x+3) = 0 }}}

Solutions (using equation of the straight line to find y values):
x = -1/2   --->  y = -1/2-4 = -1/2-8/2 = -9/2
x = -3     --->  y = -3-4 = -7

(You can verify the y values are indeed the same if you use the equation of the curve instead)


(-1/2, -9/2) and (-3,-7)  are the two points of intersection


{{{ graph(500,500,-5,5,-10,10, 0, x-4, 2x^2+8x-1) }}}