Question

on a graph of 2x^2 -7+5 use equations to find the vertex point ,roots and asis of symmetry

What the heck do i do?????

Answer 1

for starters ... you seem to be missing an x term

the vertex ("nose" of the parabola) is on the axis of symmetry

for the general expression ... a x^2 + b x + c
... the equation for the axis of symmetry is ... x = -b / (2 a)
... the axis of symmetry crosses the parabola at the vertex

the roots (where the parabola crosses the x-axis) can be found by
... setting the expression equal to zero , and using the quadratic equation
... they can also be found by factoring the expression
... if r is a root , then x-r is a factor

Answer 2

First of all you don't even have an equation and probably have a typo, I will assume it was
y = 2x^2 - 7x + 5

The vertex is the lowest point on your parabola. It can be found using
several methods:
. completing the square if you don't know Calculus
. using derivatives if you know Calculus
. short-cut method, if y = ax^2 + bx + c, the x of the vertex is -b/(2a)
using the last method, x of the vertex is 7/4
sub that back into the equation,
y = 2(49/16) - 7(7/4) + 5 = -9/8
so the vertex is (7/4, -9/8)

the roots are the points where parabola cuts across the x-axis
that is, when 2x^2 - 7x + 5 = 0
use the quadratic formula to find the two x's

by the "asis of symmetry" you meant the axis of symmetry.
for a standard parabola of your kind, that would simple be the vertical line running through the vertex, which would be x = 7/4