Question

Y=-x^2+8-12 график фу

Answer 1

Парабола будет графиком функции

Answer 2

To graph the function $y = -x^2 + 8x - 12$, we need to plot points on a coordinate plane and then connect them to form the graph. Here's how you can do it step-by-step:

Step 1: Find the vertex of the parabola. The vertex of the parabola represented by the equation $y = ax^2 + bx + c$ is given by $(h, k)$, where $h = -\frac{b}{2a}$ and $k$ is the value of the function at $x = h$.

For our equation $y = -x^2 + 8x - 12$: $a = -1$, $b = 8$, and $c = -12$.

$h = -\frac{8}{2(-1)} = 4$ and $k = -(4)^2 + 8(4) - 12 = -4$.

So, the vertex is at $(4, -4)$.

Step 2: Find the $y$-intercept. The $y$-intercept is the value of the function when $x = 0$. So, set $x = 0$ in the equation and solve for $y$.

$y = -(0)^2 + 8(0) - 12 = -12$.

So, the $y$-intercept is at $(0, -12)$.

Step 3: Find the $x$-intercepts. The $x$-intercepts are the points where the graph crosses the $x$-axis. To find these points, set $y = 0$ in the equation and solve for $x$.

$0 = -x^2 + 8x - 12$.

To solve this quadratic equation, you can use factoring or the quadratic formula. In this case, it factors nicely:

$0 = -(x^2 - 8x + 12) = -(x - 2)(x - 6)$.

Setting each factor to zero gives us two solutions:

$x - 2 = 0 \implies x = 2$ and $x - 6 = 0 \implies x = 6$.

So, the $x$-intercepts are at $(2, 0)$ and $(6, 0)$.

Step 4: Plot the points. Now that we have the vertex and the intercepts, let's plot them on the coordinate plane:

• Vertex: $(4, -4)$
• $y$-intercept: $(0, -12)$
• $x$-intercepts: $(2, 0)$ and $(6, 0)$

Step 5: Draw the graph. Now, connect the plotted points to form the graph of the function. Since it's a downward-opening parabola, the graph will look like this:

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The graph is a downward-opening parabola with the vertex at $(4, -4)$, passing through the points $(2, 0)$ and $(6, 0)$, and intersecting the $y$-axis at $(0, -12)$.

Please note that this is just a rough sketch; for a more accurate graph, you can plot more points or use graphing software.