How do you find the important points to graph $y = - x^{2} + 3$ $y = -x^2 + 3$ ?

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Answer 1 · 2018-07-02T05:45:43

vertex $(0, 3)$ $(0, 3)$
$x$ $x$ -intercepts: $(- \sqrt{3}, 0), (\sqrt{3}, 0)$ $(-sqrt(3), 0), (sqrt(3), 0)$
$y$ $y$ -intercept: $(0, 3)$ $(0, 3)$

Given: $y = - x^{2} + 3$ $y = -x^2 + 3$

With the equation in the form: $A x^{2} + B x + C = 0$ $Ax^2 + Bx + C = 0$ ,

the vertex is at $(- \frac{B}{2 A}, f (- \frac{B}{2 A}))$ $(-B/(2A), f(-B/(2A)))$ ,

the axis of symmetry is $x = - \frac{B}{2 A}$ $x = -B/(2A)$

If the coefficient $A < 0$ $A < 0$ , the vertex is a maximum

If the coefficient $A > 0$ $A > 0$ , the vertex is a minimum

For the given equation:

$- \frac{B}{2 A} = \frac{0}{- 2} = 0$ $-B/(2A) = 0/-2 = 0$

$f (0) = - {(0)}^{2} + 3 = 3$ $f(0) = -(0)^2 + 3 = 3$

vertex $(0, 3)$ $(0, 3)$ is a maximum; axis of symmetry: $x = 0$ $x = 0$

$Find x-intercepts$ $color(blue) ("Find x-intercepts")$ by setting $y = 0$ $y = 0$ :

$0 = - x^{2} + 3$ $0 = -x^2 + 3$

$- 3 = - x^{2}$ $-3 = -x^2$

$x^{2} = 3 \Rightarrow x = \pm \sqrt{3}$ $x^2 = 3 => x = +- sqrt(3)$

$x$ $x$ -intercepts: $(- \sqrt{3}, 0), (\sqrt{3}, 0)$ $(-sqrt(3), 0), (sqrt(3), 0)$

$Find y-intercept$ $color(red) ("Find y-intercept")$ by setting $x = 0$ $x = 0$ :

$y = - {(0)}^{2} + 3 \Rightarrow y = 3$ $y = -(0)^2 + 3 => y = 3$

$y$ $y$ -intercept: $(0, 3)$ $(0, 3)$ , which is the vertex

How do you find the important points to graph y = -x^2 + 3y=−x2+3y = -x^2 + 3?