f(x)=-x^2+2
the maximum y-value is y=
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x^2+14x=-8
what value should be added to both sides of the equation? the answer should be an integer or simplified fraction.
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y=x^2+6x-2
use the quadratic formula tofind any x intercepts.
To find the maximum y-value of the function f(x) = -x^2 + 2, we can identify the vertex of the parabola. The vertex of a parabola in the form y = ax^2 + bx + c can be found by using the formula x = -b/2a.
In this case, a = -1 and b = 0.
So, x = -0/2(-1) = 0.
Now, substitute this value of x back into the equation f(x) to find the corresponding y-value:
f(0) = -(0)^2 + 2 = 2.
Therefore, the maximum y-value is y = 2.
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To solve the equation x^2 + 14x = -8, we need to isolate the variable x.
To do this, we can add 8 to both sides of the equation:
x^2 + 14x + 8 = 0.
The resulting equation is quadratic and can be factored or solved using the quadratic formula, depending on its form.
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To find the x-intercepts of the quadratic equation y = x^2 + 6x - 2, we can use the quadratic formula, which is:
x = (-b ± √(b^2 - 4ac)) / (2a).
In this equation, a = 1, b = 6, and c = -2.
Substituting these values into the quadratic formula, we get:
x = (-(6) ± √((6)^2 - 4(1)(-2))) / (2(1)).
Simplifying the expression inside the square root, we have:
x = (-6 ± √(36 + 8)) / 2.
This simplifies to:
x = (-6 ± √44) / 2.
Further simplification yields:
x = (-6 ± 2√11) / 2.
Now, we can simplify the expression by canceling out the common factor of 2:
x = -3 ± √11.
So, the x-intercepts of the quadratic equation y = x^2 + 6x - 2 are x = -3 + √11 and x = -3 - √11.