A vertical parabola has vertex (1,2) and y-intercept -3. What are its x-intercepts?

Question

amistre64 · Answer

depends on if its open up or down

amistre64 · Answer

if its up; no x ints

amistre64 · Answer

but it opens down cause we got -3 as the y int lol

anonymous · Answer

vertex is (1,2) and y - intercept 3 looks like 
$$y=a(x-1)^2 +2$$

anonymous · Answer

you know that if x = 0 you get -3 so put x = 0 and solve for a

anonymous · Answer

-3=a(0-1)^2+2\]
$$-3=a +2$$
$$-5=a$$

amistre64 · Answer

-(5x^2 -10x +3) is what I got

anonymous · Answer

so your equation is $$y=-5(x-1)^2+2$$

anonymous · Answer

expand to get probably what amistre (6 to go!) got

amistre64 · Answer

id rather keep it your way and get the sqrts

anonymous · Answer

actually i think it is $$-5x^2+10x-3$$

amistre64 · Answer

it is, but the negative 1 factors out to make life easier

anonymous · Answer

sorry

anonymous · Answer

ok i wasn't paying attention you are right!

amistre64 · Answer

0=−5(x−1)^2+2

5(x−1)^2 = 2

(x−1)^2 = 2/5

x−1 = +- sqrt(2/5)

x = 1  +- sqrt(2/5)

amistre64 · Answer

your equation is quicker to find the xints from :)

anonymous · Answer

now set = 0 and solve
$$-5x^2+10x-3=0$$
$$5x^2-10x+3=0$$

anonymous · Answer

that is for sure.  should have left it that way!

amistre64 · Answer

should have lol

anonymous · Answer

standard form might be 
$$1\pm \frac{\sqrt{10}}{5}$$

anonymous · Answer

or even $$\frac{5\pm\sqrt{10}}{5}$$