Question 177856
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We know the standard eqn of a parabola: {{{y=ax^2+bx+c}}}
Being the vertex form---->{{{y=a(x-h)^2+k}}}, where (h,k) is the vertex:
I'll do the first one, and you can continue the rest;
33. {{{y = x2 + 4x - 7}}}--->follows std eqn, where{{{system(a=1,b=4,c=-7)}}}
Complete the square, adding a constant by taking half of the "b" constant then squared. In this case the "b" constant is {{{4}}}, half of it {{{4/2=2}}} then squared, {{{2^2=highlight(4)}}}:
{{{y=(x^2+4x+highlight(4))-7-highlight(4)}}}, Also subtract what you added so the process won't change.
{{{highlight(y=(x+2)^2-11)}}}---->it follows the vertex form, being {{{system(h=x=-2,k=y=-11)}}}
To get the x-intercept, we solve the eqn by Quadratic:
where---->{{{system(a=1,b=4,c-7)}}}
{{{x=(-b+-sqrt(b^2-4ac))/(2a)=(-4+-sqrt(4^2-4*1*-7))/(2*1)}}}
{{{x=(-4+-sqrt(16+28))/2=(-4+-sqrt(44))/2=(-4+-6.63)/2}}}
{{{x=(-4+6.63)/2=2.63/2=highlight(1.32=x)}}}
{{{x=(-4-6.63)/2=-10.63/2=highlight(-5.32=x)}}}
For the Y-Intercept, let {{{f(x)=0}}}
{{{y=0^2+4*0-7=highlight(-7=y)}}}
As we see the graph:
{{{drawing(400,400,-8,8,-15,10,grid(1),graph(400,400,-8,8,-15,10,x^2+4x-7),blue(circle(-2,-11,.25)),green(circle(1.32,0,.20)),green(circle(-5.32,0,.20)),green(circle(0,-7,.20)))}}}
 Thank you,
Jojo</pre>