-1
f(x)=x^2
(y - 3) = a(x - 1)2 y = a(x - 1)2 + 3 4 = a(4 - 1)2 + 3 1 = 9a a = 1/9 y = 1/9 (x - 1)2 + 3
Assuming the vertex is 0,0 and the directrix is y=4 x^2=0
-3x2+12x-1 = y Solve the quadratic equation when y = 0 by means of the quadratic equation formula which gives x values of 3.914854216 and 0.085145784. Add these values together and divide them by 2 which is 4/2 = 2 and this is the line symmetry of the parabola. Substitute 2 for x into the original equation to find the value of y: So the vertex is at (2, 11) Remember that the parabola has a maximum value because the coefficient of x2 is negative in other words it will face downwards.
5
-2
-1
The vertex of this parabola is at -2 -3 When the y-value is -2 the x-value is -5. The coefficient of the squared term in the parabola's equation is -3.
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)
2
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) = (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
Y=a(x-h)+k is the vertex formula. Since the vertex is at (-2,-3) this parabola has the equation: y=a(x+2)^2-3 We can plug in x=-1 but we really need to know a, to solve for y. ( we can solve it, but we will have an a in the solution)
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The coordinates will be at the point of the turn the parabola which is its vertex.
f(x)=x^2