1. Find the equation for a quadratic function whose vertex is (2,5) and whose graph contains the point (-8,15).
2.What are the x- and y- intercepts for y=2x^2+x-6
Plesee explai. I want to learn to learn how to do them.
The standard form of a quadratic is
y = a(x-h)^2 + k, where (h,k) is the vertex.
since we know the vertex we can start with
y = a(x-2)^2 + 5 , leaving us only with the a as an unknown, but we know the point (-8,15) lies on it, thus must satisfy our equation, so ...
15 = a(-8-2)^2 + 5
10 = 100a
a = 10/100 = 1/10
equation: y = (1/10)(x-2)^2 + 5
2.
for the y-intercept, let x = 0
we can visually see the would give us y = 6
for the x-intercept, let y = 0
then
2x^2 + x - 6 = 0
(2x - 3)(x + 2) = 0
x = 3/2 or x = -2
1. To find the equation of a quadratic function given the vertex and a point on the graph, we can use the vertex form of the quadratic equation: y = a(x - h)^2 + k, where (h, k) represents the vertex.
Given that the vertex is (2,5), we have h = 2 and k = 5. Plugging these values into the equation, we have:
y = a(x - 2)^2 + 5
Now, we can use the given point (-8,15) to find the value of 'a'. Plugging these values into the equation, we get:
15 = a(-8 - 2)^2 + 5
15 = a(-10)^2 + 5
15 = 100a + 5
100a = 15 - 5
100a = 10
a = 10/100
a = 0.1
Therefore, the equation for the quadratic function is y = 0.1(x - 2)^2 + 5.
2. To find the x-intercepts of a quadratic equation, we set y = 0 and solve for x.
Given the equation y = 2x^2 + x - 6, we set y = 0:
0 = 2x^2 + x - 6
To solve this quadratic equation, we can factor or use the quadratic formula. In this case, factoring will work:
0 = (2x - 3)(x + 2)
Setting each factor equal to zero:
2x - 3 = 0
x + 2 = 0
Solving for x, we get:
2x = 3
x = 3/2 = 1.5
x = -2
Therefore, the x-intercepts are x = 1.5 and x = -2.
To find the y-intercept, we set x = 0:
y = 2(0)^2 + (0) - 6
y = 0 - 6
y = -6
Therefore, the y-intercept is y = -6.
To find the equation for a quadratic function given its vertex and a point on its graph, you can use the standard form of a quadratic equation: y = ax^2 + bx + c.
1. Finding the equation for a quadratic function:
Step 1: Substitute the coordinates of the vertex into the equation to find the values of 'a', 'b', and 'c'.
When the vertex is (h, k), the equation becomes: y = a(x - h)^2 + k.
In this case, the vertex is (2, 5), so the equation becomes: y = a(x - 2)^2 + 5.
Step 2: Substitute the given point (-8, 15) into the equation to find the value of 'a'.
Replace 'x' and 'y' with the coordinates of the point and solve for 'a':
15 = a(-8 - 2)^2 + 5
Simplifying the equation:
15 = 100a + 5
10 = 100a
Dividing both sides by 100:
a = 10/100
a = 0.1
Step 3: Substitute 'a' back into the equation to get the final equation:
y = 0.1(x - 2)^2 + 5
Therefore, the equation for the quadratic function with the given vertex (2, 5) and the point (-8, 15) is y = 0.1(x - 2)^2 + 5.
2. Finding the x- and y-intercepts of y=2x^2+x-6:
The x-intercepts are the points where the graph intersects the x-axis. To find them, set y = 0 and solve for x.
Setting y = 0 in the equation:
0 = 2x^2 + x - 6
This quadratic equation can be factored or solved using the quadratic formula. In this case, it can be factored as (2x - 3)(x + 2) = 0.
Setting each factor equal to zero:
2x - 3 = 0
x + 2 = 0
Solving for x:
2x = 3 --> x = 3/2 or 1.5
x = -2
Therefore, the x-intercepts are (1.5, 0) and (-2, 0).
The y-intercept is the point where the graph intersects the y-axis. To find it, set x = 0 in the equation.
Setting x = 0 in the equation:
y = 2(0)^2 + 0 - 6
Simplifying the equation:
y = -6
Therefore, the y-intercept is (0, -6).
By following these steps, you can find the equation for a quadratic function given its vertex and a point on its graph, as well as determine the x- and y-intercepts of a quadratic equation.