Question 184671
{{{y = 3x^2 + (k+2)x + 3}}}
Instead of 2 roots, {{{r[1]}}} and {{{r[2]}}},
There will be a double root, {{{r}}}
{{{(x - r)(x - r) = 0}}}
{{{x^2 - 2rx + r^2 = 0}}}
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{{{3x^2 + (k+2)x + 3 = 0}}}
{{{x^2 + ((k+2)/3)*x + 1 = 0}}}
I can make the substitutions
{{{(k+2)/3 = -2r}}}
{{{r^2 = 1}}}
{{{r = 1}}}
and
{{{r = -1}}}
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{{{(k+2)/3 = -2*1}}}
{{{k + 2 = -6}}}
{{{k = -8}}}
and
{{{(k+2)/3 = -2*(-1)}}}
{{{k + 2 = 6}}}
{{{k = 4}}}
The graph touches the x-axis when {{{k = -8}}} or {{{k = 4}}}
for {{{k = -8}}}:
{{{y = 3x^2 + (k+2)x + 3}}}
{{{y = 3x^2 + (-8+2)x + 3}}}
{{{y = 3x^2 - 6x + 3}}}
for {{{k = 4}}}:
{{{y = 3x^2 + (4+2)x + 3}}}
{{{y = 3x^2 + 6x + 3}}}
I'll plot both:
 {{{ graph( 500, 500, -5, 5, -2, 8, 3x^2 - 6x + 3,3x^2 + 6x + 3) }}}