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The equation of a parabola with its vertex at the point (-36, k) can be expressed in the vertex form as ( y = a(x + 36)^2 + k ), where ( a ) determines the direction and width of the parabola. If the vertex is at (-36), the x-coordinate is fixed, but the y-coordinate ( k ) can vary depending on the specific position of the vertex. If you'd like a specific example, assuming ( k = 0 ) and ( a = 1 ), the equation would be ( y = (x + 36)^2 ).
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To find the equation of a parabola with vertex at ((-3, 0)) that passes through the point ((3, 18)), we can use the vertex form of a parabola, (y = a(x + 3)^2). To determine the value of (a), substitute the point ((3, 18)) into the equation: [ 18 = a(3 + 3)^2 \implies 18 = a(6)^2 \implies 18 = 36a \implies a = \frac{1}{2}. ] Thus, the equation of the parabola is (y = \frac{1}{2}(x + 3)^2).
Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
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Go study
The vertex of this parabola is at 5 5 When the x-value is 6 the y-value is -1. The coefficient of the squared expression in the parabola's equation is -6.
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Since the vertex is at the origin and the parabola opens downward, the equation of the parabola is x2 = 4py, where p < 0, and the axis of symmetry is the y-axis. So the focus is at y-axis at (0, p) and the directrix equation is y = -p. Now, what do you mean with 1 and 76 units? 1.76 units? If the distance of the vertex and the focus is 1.76 units, then p = -1.76, thus 4p = -7.04, then the equation of the parabola is x2 = -7.04y.
The vertex of this parabola is at -3 -1 When the y-value is 0 the x-value is 4. The coefficient of the squared term in the parabolas equation is 7
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (2, -1), and a point on it is (5, 0) → 0 = a(5 - 2)² + -1 → 0 = a(3)² -1 → 1 = 9a → a = 1/9 → The coefficient of the x² term is 1/9
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (-2, -3), and a point on it is (-1, -5) → -5 = a(-1 - -2)² + -3 → -5 = a(1)² - 3 → -5 = a - 3 → a = -2 → The coefficient of the x² term is -2.
A parabola with vertex (h, k) has equation of the form: y = a(x - h)² + k → vertex (k, h) = (3, 5), and a point on it is (-1, 6) → 6 = a(-1 - 3)² + 5 → 6 = a(-4)² + 5 → 1 = 16a → a = 1/16 → The coefficient of the x² term is 1/16
the vertex of a parabola is the 2 x-intercepts times-ed and then divided by two (if there is only 1 x-intercept then that is the vertex)
The vertex of a parabola is found by using the solution of the equation -b/2a and putting it into the quadratic equation. a is the coefficient of x^2. b is the coefficient of the other x in the equation. Ex. y=2x^2+2x+1 -b/2a = -2/2(2) = -1/2 Now put -1/2 in the place of every x in the equation. y=2(-1/2)^2+2(-1/2)+1 The vertex is (-1/2, 1/2)