Answer:
Zeros:
[tex]x_1=-1+\sqrt{2}[/tex]
[tex]x_2=-1-\sqrt{2}[/tex]
Interception with y-axis
[tex]y=- 1[/tex]
Vertex:
(-1, -2)
Step-by-step explanation:
We have the following quadratic function
[tex]f(x) = x^2 + 2x - 1[/tex]
To find the zeros of the function, make [tex]f(x) = 0[/tex] and solve for the variable x
[tex]x^2 + 2x - 1=0[/tex]
We must factor the expression. Then we use the quadratic formula:
For a function of the form [tex]ax ^ 2 + bx + c = 0[/tex] the quadratic formula is:
[tex]x=\frac{-b\±\sqrt{b^2-4ac}}{2a}[/tex]
In this case note that:
[tex]a=1\\b=2\\c=-1[/tex]
Then:
[tex]x=\frac{-2\±\sqrt{2^2-4(1)(-1)}}{2}[/tex]
[tex]x=\frac{-2\±\sqrt{8}}{2}[/tex]
[tex]x=\frac{-2\±2\sqrt{2}}{2}[/tex]
[tex]x_1=-1+\sqrt{2}[/tex]
[tex]x_2=-1-\sqrt{2}[/tex]
We know that the vertex of a quadratic function is at the point:
[tex](-\frac{b}{2a}, f(-\frac{b}{2a}))[/tex]
Then:
[tex]x=-\frac{2}{2*1}=-1[/tex]
[tex]y=f(-1) = (-1)^2+2(-1) -1 = -2[/tex]
The vertex is: (-1, -2)
The intersection with the y-axis we find it doing [tex]x = 0[/tex] and solving for y
[tex]y=f(0) = 0^2 + 2*0 - 1[/tex]
[tex]y=- 1[/tex]
Look at the attached image.
