```Question 177821
</pre><font size=4><b>
a) {{{y=x^2-4x+4}}}
Factorable being a perfect square: {{{(x-2)(x-2)=0}}}
it shows you got X-intercept, (2,0)
Let' see the vertex to see where we are, via Vertex Form: {{{y=a(x-h)^2+k2}}}
We complete the square of the given eqn:
{{{(x^2-4x+4)+4-4}}}
{{{(x-2)^2+0}}}
The vertex exists @ (2,0)
Letting {{{f(x)=0}}}
{{{y=0^2-4(0)+4}}}
{{{y=4}}}, Y-Intercept
Conclusion, given eqn it has ONLY X-Intercept because the vertex lies on the x-axis.
As you see below;
{{{drawing(300,300,-6,6,-6,6,grid(1),graph(300,300,-6,6,-6,6,x^2-4x+4),green(circle(0,4,.16)),blue(circle(2,0,.16)),blue(circle(2,0,.12)))}}}
b){{{y=4x^2-12x+9}}}
For this solution toe xist, the discriminant should not be a negative number:
By Quadratic so we can see the roots (X-intercepts):
where{{{system(a=4,b=-12,c=9)}}}
Solving for the discriminant: {{{-b^2-4ac=-12^2-4*4*0=144-144=0}}}
The dicriminant is zero. Therefore, there's only one solution.
{{{x=(-b+-sqrt(b^2-4ac))/(2a)=(-(-12)+-sqrt(-12^2-4*4*9))/(2*4)}}}
{{{x=(12+-sqrt(144-144))/8=(12+-0)/8=12/8=highlight(1.5=x)}}}
Let {{{f(x)=0}}}
{{{y=4(0)^2-12(0)+9}}}
{{{y=9}}} Y-intercept
{{{drawing(300,300,-3,5,-3,10,grid(1),graph(300,300,-3,5,-3,10,4x^2-12x+9),blue(circle(1.5,0,.16)),blue(circle(1.5,0,.12)),green(circle(0,9,.16)))}}}
Thank you,
Jojo</pre>
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