The given equation is a quadratic function in the form of y = ax^2 + bx + c.
Comparing the given equation with the standard form, we can see that a = -2, b = 3, and c = -9.
To find the vertex of the parabola, we can use the vertex formula, which is x = -b / (2a).
Substituting the values a = -2 and b = 3 into the formula, we get x = -3 / (2 * -2) = -3 / -4 = 3/4.
To find the y-coordinate of the vertex, substitute the x-coordinate (3/4) back into the given equation:
y = -2(3/4)^2 + 3(3/4) - 9
y = -2(9/16) + 9/4 - 9
y = -18/16 + 36/16 - 144/16
y = -126/16
y = -63/8
Therefore, the vertex of the parabola is (3/4, -63/8).
y= -2x^2 + 3x - 9
11 answers
The graph of a parabola passes through the points (-4,8) and (2,8). The equation of the axis of symmetry of the graph of the quadratic function is
To find the equation of the axis of symmetry of a parabola, we need to find the x-coordinate of the vertex. The x-coordinate of the vertex is the average of the x-values of the two given points (-4,8) and (2,8), since the axis of symmetry passes through the vertex and is equidistant from the two points.
(x₁ + x₂) / 2 = (-4 + 2) / 2 = -2 / 2 = -1
Therefore, the x-coordinate of the vertex is -1.
The equation of the axis of symmetry is x = -1.
(x₁ + x₂) / 2 = (-4 + 2) / 2 = -2 / 2 = -1
Therefore, the x-coordinate of the vertex is -1.
The equation of the axis of symmetry is x = -1.
The x-intercepts of the parabola represented by the function are
f(x) = 3x^2 - 9x - 30
f(x) = 3x^2 - 9x - 30
To find the x-intercepts of the parabola represented by the function f(x) = 3x^2 - 9x - 30, we need to set the function equal to zero and solve for x.
So, we have:
3x^2 - 9x - 30 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's try factoring first.
We need to find two numbers that multiply to -90 (the product of the coefficient of x^2 and the constant term) and add up to -9 (the coefficient of x). The numbers that satisfy these conditions are -15 and 6.
So, we can rewrite the equation as:
3x^2 - 15x + 6x - 30 = 0
Now, let's factor by grouping:
(3x^2 - 15x) + (6x - 30) = 0
3x(x - 5) + 6(x - 5) = 0
(3x + 6)(x - 5) = 0
Now, we can set each factor equal to zero and solve for x:
3x + 6 = 0 or x - 5 = 0
For the first equation:
3x + 6 = 0
3x = -6
x = -6/3
x = -2
For the second equation:
x - 5 = 0
x = 5
Therefore, the x-intercepts of the parabola are x = -2 and x = 5.
So, we have:
3x^2 - 9x - 30 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's try factoring first.
We need to find two numbers that multiply to -90 (the product of the coefficient of x^2 and the constant term) and add up to -9 (the coefficient of x). The numbers that satisfy these conditions are -15 and 6.
So, we can rewrite the equation as:
3x^2 - 15x + 6x - 30 = 0
Now, let's factor by grouping:
(3x^2 - 15x) + (6x - 30) = 0
3x(x - 5) + 6(x - 5) = 0
(3x + 6)(x - 5) = 0
Now, we can set each factor equal to zero and solve for x:
3x + 6 = 0 or x - 5 = 0
For the first equation:
3x + 6 = 0
3x = -6
x = -6/3
x = -2
For the second equation:
x - 5 = 0
x = 5
Therefore, the x-intercepts of the parabola are x = -2 and x = 5.
Determine the vertex of the parabola represented by the function
y = -8(x+3)^2 + 5
y = -8(x+3)^2 + 5
The given equation is already in vertex form: y = a(x - h)^2 + k, where (h, k) represents the vertex of the parabola.
Comparing the given equation with the vertex form, we can see that h = -3 and k = 5.
Therefore, the vertex of the parabola is (-3, 5).
Comparing the given equation with the vertex form, we can see that h = -3 and k = 5.
Therefore, the vertex of the parabola is (-3, 5).
Determine the standard form of the function f(x) = -2(x-3)^2 + 4
To convert the given function f(x) = -2(x-3)^2 + 4 into standard form, we need to expand and simplify the equation.
Let's start by expanding the square term (x-3)^2:
f(x) = -2(x^2 - 6x + 9) + 4
Next, distribute the -2 to each term inside the parentheses:
f(x) = -2x^2 + 12x - 18 + 4
Combine like terms:
f(x) = -2x^2 + 12x - 14
Therefore, the standard form of the function f(x) = -2(x-3)^2 + 4 is f(x) = -2x^2 + 12x - 14.
Let's start by expanding the square term (x-3)^2:
f(x) = -2(x^2 - 6x + 9) + 4
Next, distribute the -2 to each term inside the parentheses:
f(x) = -2x^2 + 12x - 18 + 4
Combine like terms:
f(x) = -2x^2 + 12x - 14
Therefore, the standard form of the function f(x) = -2(x-3)^2 + 4 is f(x) = -2x^2 + 12x - 14.
An explosives company specializes in bringing down old buildings. The time it takes for a particular piece of flying debris to fall to the ground can be modelled by the function h(t) = -18t^2 + 72t + 378, where h is the height of the debris, in metres, and t is the time, in seconds.
How long will it take for the piece of debris to hit the ground after an explosion?
How long will it take for the piece of debris to hit the ground after an explosion?
To find the time it takes for the debris to hit the ground, we need to find the value of t when h(t) = 0.
The given function is h(t) = -18t^2 + 72t + 378.
Setting h(t) = 0:
-18t^2 + 72t + 378 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -18, b = 72, and c = 378.
Substituting these values into the quadratic formula, we have:
t = (-72 ± √(72^2 - 4(-18)(378))) / (2(-18))
t = (-72 ± √(5184 + 27312)) / (-36)
t = (-72 ± √(32496)) / (-36)
t = (-72 ± 180) / (-36)
We have two possible solutions:
t₁ = (-72 + 180) / (-36)
t₂ = (-72 - 180) / (-36)
Calculating these values:
t₁ = 108 / (-36)
t₁ = -3
t₂ = -252 / (-36)
t₂ = 7
Since time cannot be negative in this context, the time it takes for the debris to hit the ground after an explosion is t = 7 seconds.
The given function is h(t) = -18t^2 + 72t + 378.
Setting h(t) = 0:
-18t^2 + 72t + 378 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -18, b = 72, and c = 378.
Substituting these values into the quadratic formula, we have:
t = (-72 ± √(72^2 - 4(-18)(378))) / (2(-18))
t = (-72 ± √(5184 + 27312)) / (-36)
t = (-72 ± √(32496)) / (-36)
t = (-72 ± 180) / (-36)
We have two possible solutions:
t₁ = (-72 + 180) / (-36)
t₂ = (-72 - 180) / (-36)
Calculating these values:
t₁ = 108 / (-36)
t₁ = -3
t₂ = -252 / (-36)
t₂ = 7
Since time cannot be negative in this context, the time it takes for the debris to hit the ground after an explosion is t = 7 seconds.