HELP! 18 points for the two questions!! urgent will mark brainliest

2- yes the parabola opens upward but none are right in the right mix for the other two. the vertex is at (0,-5) and the axis of symmetry is x = 0 (not only because I calculated it and the graph showed it but because (x,y) And the x spot is a 0)

any time you just see a plain x^2 or really anything like that in the same vanilla sense wether the exponent is changed or there is a coefficient and all it has is a plus or minus whatever it will just be along the y axis where the plus or minus number was. (can't remember any vanilla advice for parabolas sorry)

kudzordzifrancis kudzordzifrancis · Answer 2 · 2019-05-26T10:04:17+02:00

Answer:

1. a=5,h=0,k=0

2. None of the choices are correct

Step-by-step explanation:

1. The function given is:[tex]f(x)=5x^2[/tex]

We can rewrite this in vertex form as: [tex]f(x)=5(x-0)^2+0[/tex].

Comparing this to [tex]f(x)=a(x-h)^2+k[/tex], we have a=5,h=0 and k=0.

The first choice is correct

2. The function given is: [tex]f(x)=x^2-5[/tex]

This can be rewritten in the vertex form as:

[tex]f(x)=a(x-0)^2-5[/tex]

Comparing this to [tex]f(x)=a(x-h)^2+k[/tex], we have a=1,h=0 and k=-5.

Since 'a' is positive, the parabola opens upwards, and has its vertex at (h,k)=(0,-5), and line of symmetry at x=0.

Based on this, the correct choice is none of the choices are correct