What is the vertex of y=7(x+5)^2+4?
well, surely x = -5
then
y = 4
Oh, the vertex of that quadratic equation is where all the fun happens! So, let's find it together. We have y = 7(x+5)^2 + 4.
First, we can see that the expression inside the parentheses, (x+5), is a quadratic term. To find the vertex, we need to make it look like a perfect square.
So, let's expand it: y = 7(x^2 + 10x + 25) + 4.
Now, we can distribute the 7: y = 7x^2 + 70x + 175 + 4.
Combining like terms, we have: y = 7x^2 + 70x + 179.
Now, let's take a look at the coefficient of x^2, which is 7. Since it's positive, we know that the parabola opens upward. Therefore, the vertex is the lowest point on the graph.
To find the x-coordinate of the vertex, we use the formula: x = -b / (2a). In this case, a = 7 and b = 70.
Plugging in the values, we get x = -70 / (2 * 7), which simplifies to x = -5.
We know the x-coordinate, but what about the y-coordinate? Well, we can substitute x = -5 into the equation to find it.
When x = -5, y = 7(-5)^2 + 70(-5) + 179. Simplifying the expression, we get y = 7(25) - 350 + 179.
Calculating further, y = 175 - 350 + 179, which gives us y = 4.
So, after all that comedy, the vertex of the quadratic equation y = 7(x+5)^2 + 4 is (-5, 4). Now, isn't that a hilarious place to be?
To find the vertex of the quadratic equation given by y = 7(x + 5)^2 + 4, we can use the vertex form of a quadratic equation, which is y = a(x - h)^2 + k. In this equation, the vertex is represented by the coordinates (h, k).
Comparing the given equation with the vertex form, we have:
a = 7
h = -5
k = 4
So, the vertex of the equation y = 7(x + 5)^2 + 4 is (-5, 4).
To find the vertex of a quadratic function in the form y = a(x - h)^2 + k, where (h, k) represents the vertex, you can use the formula:
h = -b/2a
In your equation, y = 7(x+5)^2 + 4, we can identify a = 7, h = -5 (opposite sign of the constant term inside the parentheses), and k = 4.
Using the formula, we can calculate the x-coordinate of the vertex as follows:
h = -(-5)/(2 * 7)
= 5/14
So, the x-coordinate of the vertex is 5/14.
To find the y-coordinate of the vertex, substitute this x-coordinate value (-5/14) back into the original equation:
y = 7(-5/14 + 5)^2 + 4
Simplify the expression inside parentheses:
y = 7(-5/14 + 70/14) + 4
= 7(65/14) + 4
Combine the fractions:
y = (455/14) + (56/14)
= 511/14
So, the y-coordinate of the vertex is 511/14.
Therefore, the vertex of the quadratic function y = 7(x+5)^2 + 4 is approximately (5/14, 511/14).