Please help!
Find the vertex of f(x)=x^2+2x-3
Find the vertex of f(x)=x^2-3x+1
the vertex is on the axis of symmetry
... x = -b / 2a
f(x)=x^2+2x-3
... x = -2 / 2*1 = -1
f(x)=x^2-3x+1
... x = - -3 / 2*1 = 3/2
plug the x's into their functions to find the y's for the vertex coordinates
Thank you!
To find the vertex of a quadratic function in the form f(x) = ax^2 + bx + c, you can use the formula x = -b / (2a) to find the x-coordinate of the vertex. Then, substitute this x-value into the original equation to find the y-coordinate.
1. Let's first find the vertex of f(x) = x^2 + 2x - 3:
a = 1, b = 2, and c = -3.
We can use the formula x = -b / (2a).
Substituting the values, we have x = -2 / (2*1) = -2 / 2 = -1.
Now, substitute x = -1 into the original equation f(x) = x^2 + 2x - 3:
f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4.
So, the vertex of f(x) = x^2 + 2x - 3 is (-1, -4).
2. Now let's find the vertex of f(x) = x^2 - 3x + 1:
a = 1, b = -3, and c = 1.
Again, we can use the formula x = -b / (2a).
Substituting the values, we have x = -(-3) / (2*1) = 3 / 2 = 1.5.
Now substitute x = 1.5 into the original equation f(x) = x^2 - 3x + 1:
f(1.5) = (1.5)^2 - 3(1.5) + 1 = 2.25 - 4.5 + 1 = -1.25.
So, the vertex of f(x) = x^2 - 3x + 1 is (1.5, -1.25).
Therefore, the vertex of f(x) = x^2 + 2x - 3 is (-1, -4), and the vertex of f(x) = x^2 - 3x + 1 is (1.5, -1.25).