Answer:
Vertex and y-intercept (0, -5)
x-intercepts (-1.58, 0) (1.58, 0)
opens upwards
other plot points: (-2, 3) (-1, -3) (1, -3) (2, 3)
Step-by-step explanation:
The graph is not quite correct - it's a little too narrow and doesn't go through the points on the graph.
The y-intercept is when x = 0:
f(0) = 2(0)² - 5
= - 5
Therefore, the y-intercept is at (0, -5)
We also know that the y-intercept is the vertex since the equation is in the form [tex]f(x)=ax^2+c[/tex]
The x-intercepts are when f(x) = 0:
[tex]\implies 2x^2 - 5 = 0[/tex]
[tex]\implies x^2 =\dfrac52[/tex]
[tex]\implies x=\pm1.58113883...[/tex]
As the leading coefficient is positive, the parabola opens upwards.
Finally, input values -2 ≤ x ≤ 2 to find plot points:
(-2, 3)
(-1, -3)
(0, -5)
(1, -3)
(2, 3)
