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?????
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
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
Don't worry, I can help you with that! To find the vertex point, roots, and axis of symmetry of the given quadratic equation, 2x^2 - 7x + 5, we can use equations specific to quadratic functions.
1. Vertex Point:
The vertex of a quadratic function can be found using the formula:
x = -b / (2a)
where "a" is the coefficient of x^2 term and "b" is the coefficient of the x term in the equation. In this case, a = 2 and b = -7.
Substituting the values, we get:
x = -(-7) / (2 * 2)
x = 7 / 4
To find the y-coordinate of the vertex, substitute this x-value back into the equation:
y = 2(7/4)^2 - 7(7/4) + 5
y = 49/8 - 49/4 + 5
y = 49/8 - 98/8 + 40/8
y = -9/8
So, the vertex point is (7/4, -9/8).
2. Roots:
The roots (also called x-intercepts or zeros) are the x-values where the graph of the equation intersects the x-axis. In other words, they are the values of x for which the equation equals zero.
The equation 2x^2 - 7x + 5 can be solved when it equals zero:
2x^2 - 7x + 5 = 0
To find the roots, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -7, and c = 5.
Substituting the values, we get:
x = (-(-7) ± √((-7)^2 - 4 * 2 * 5)) / (2 * 2)
x = (7 ± √(49 - 40)) / 4
x = (7 ± √9) / 4
x = (7 ± 3) / 4
So the roots are:
x1 = (7 + 3) / 4 = 10 / 4 = 5/2
x2 = (7 - 3) / 4 = 4 / 4 = 1
Thus, the roots are x = 5/2 and x = 1.
3. Axis of Symmetry:
The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two symmetric halves. The equation of the axis of symmetry can be determined by the x-coordinate of the vertex.
For this equation, the axis of symmetry passes through x = 7/4.
So, the equation of the axis of symmetry is x = 7/4.
I hope that helps!