Determine the vertex, axis of symmetry, x-intercept, and y -intercept for the equation: y=−2x2+12x−13
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To determine the vertex, axis of symmetry, x-intercept, and y-intercept for the equation: y = -2x^2 + 12x - 13, we can start by looking at the general form of a quadratic equation, y = ax^2 + bx + c.
To find the vertex, we use the formula x = -b / (2a). So in this case, we have x = -12 / (2 * -2) = -12 / -4 = 3.
To find the y-coordinate of the vertex, we substitute the x-coordinate into the equation: y = -2(3)^2 + 12(3) - 13 = -18 + 36 - 13 = 5.
Therefore, the vertex is (3, 5).
The axis of symmetry is a vertical line that passes through the vertex. For this equation, the axis of symmetry is the line x = 3.
To find the x-intercepts, we set y = 0 and solve for x:
0 = -2x^2 + 12x - 13
We can solve this quadratic equation by factoring or using the quadratic formula. In this case, the equation does not factor easily, so we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -2, b = 12, and c = -13:
x = (-12 ± √(12^2 - 4(-2)(-13))) / (2*(-2))
= (-12 ± √(144 - 104)) / (-4)
= (-12 ± √40) / (-4)
= (-12 ± 2√10) / (-4)
= 3 ± (1/2)√10
So the x-intercepts are approximately x = 3 + (1/2)√10 and x = 3 - (1/2)√10.
To find the y-intercept, we set x = 0 and solve for y:
y = -2(0)^2 + 12(0) - 13
= 0 + 0 - 13
= -13
Therefore, the y-intercept is (0, -13).