Find the coordinates of the vertex and determine whether the graph opens up or down. y=-x^2+x-5
it opens up, because the (x^2) is positive. if it was negative, it would open down. The vertex is as follows:
to find the value of vertex, solve first for when x=o, then when y=0
Actually, note that
y=-x^2+x-5
so it opens downward
y = -(x^2 - x + 1/4) - 5 + 1/4
= -(x - 1/2)^2 - 19/4
Now it should be clear where the vertex lies.
To find the vertex and determine whether the graph of the quadratic equation y = -x^2 + x - 5 opens up or down, we can start by recognizing that the equation is in the standard form y = ax^2 + bx + c.
In this case:
a = -1
b = 1
c = -5
To find the x-coordinate of the vertex, we use the formula x = -b / (2a).
Substituting the values, we get:
x = -1 / (2 * -1) = 1 / 2
To find the y-coordinate of the vertex, we substitute the x-coordinate we just found into the equation. So:
y = - (1 / 2)^2 + (1 / 2) - 5
y = -1 / 4 + 1 / 2 - 5
y = -1 / 4 - 2 / 4 - 20 / 4
y = -23 / 4
Therefore, the coordinates of the vertex are (1 / 2, -23 / 4).
To determine whether the graph opens up or down, we look at the coefficient 'a'. If 'a' is positive, the graph opens upward; if 'a' is negative, the graph opens downward.
In this case, since 'a' is -1 (negative), the graph of y = -x^2 + x - 5 opens downward.