The question asks you to find the vertex of the function: f(x) = x^2 + 6x + 5
I have no idea how to solve this problem. Please help me understand and explain your steps! thank you!
Also: The problem is multiple choice, however I need to show my work to solve the problem so I can't just plug in the answers into the equation. I know that the answer is (-3,-4), but I don't know how to get there.
Please help!
Thank you!!
I have no idea how to solve this problem. Please help me understand and explain your steps! thank you!
Also: The problem is multiple choice, however I need to show my work to solve the problem so I can't just plug in the answers into the equation. I know that the answer is (-3,-4), but I don't know how to get there.
Please help!
Thank you!!
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Method 1:
The axis of symmetry of a parabola in the form [ax² + bx + c] occurs at [x = -b/(2a)].
x = -6/(2*1)
x = -3
Plug in for x to solve for y:
f(-3) = (-3)² + 6(-3) + 5
f(-3) = 9 - 18 + 5
f(-3) = -4
Vertex: (-3,-4)
--------------------------------------…
Method 2:
Complete the square:
f(x) = x² + 6x + 5
f(x) = (x² + 6x + 9) + 5 - 9
f(x) = (x + 3)² - 4
The parabola is now in vertex form, [y = (x - h)² + k], where (h,k) is the vertex.
Vertex: (-3, -4).
The axis of symmetry of a parabola in the form [ax² + bx + c] occurs at [x = -b/(2a)].
x = -6/(2*1)
x = -3
Plug in for x to solve for y:
f(-3) = (-3)² + 6(-3) + 5
f(-3) = 9 - 18 + 5
f(-3) = -4
Vertex: (-3,-4)
--------------------------------------…
Method 2:
Complete the square:
f(x) = x² + 6x + 5
f(x) = (x² + 6x + 9) + 5 - 9
f(x) = (x + 3)² - 4
The parabola is now in vertex form, [y = (x - h)² + k], where (h,k) is the vertex.
Vertex: (-3, -4).
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This is a quadratic equation (highest order term is x^2) so its graph will be a parabola. This curve only has one turning point, which will be its vertex.
To find it you therefore need to find where the graph is flat, i.e. has zero gradient.
That means you need to differentiate f(x) to get f'(x).
Then solve f'(x) = 0 (because we want the point where the slope's gradient is zero). To give a hint, f(x) is a quadratic, so f'(x) will be a linear function.
That gives you the value of x when the slope has zero gradient (the vertex). Plug that back into the original f(x) to get the y value of the vertex.
To find it you therefore need to find where the graph is flat, i.e. has zero gradient.
That means you need to differentiate f(x) to get f'(x).
Then solve f'(x) = 0 (because we want the point where the slope's gradient is zero). To give a hint, f(x) is a quadratic, so f'(x) will be a linear function.
That gives you the value of x when the slope has zero gradient (the vertex). Plug that back into the original f(x) to get the y value of the vertex.