Hallar las coordenadas del vértice,los puntos de corte con el eje $$ x $$ y graficar las siguiente función $$ y=-10 x^{2}+12 x+35 $$ $$ 5: 20 \text { a.m } $$

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Alright, I need to find the coordinates of the vertex, the points where the function $$ y = -10x^2 + 12x + 35 $$ intersects the x-axis, and then graph the function. Let's break this down step by step.

First, to find the vertex of a quadratic function in the form $$ y = ax^2 + bx + c $$, I can use the formula for the x-coordinate of the vertex: $$ x = -\frac{b}{2a} $$. Here, $$ a = -10 $$ and $$ b = 12 $$. Plugging these values in:

$$
x = -\frac{12}{2 \times (-10)} = -\frac{12}{-20} = 0.6
$$

Now, to find the y-coordinate of the vertex, I'll substitute $$ x = 0.6 $$ back into the original equation:

$$
y = -10(0.6)^2 + 12(0.6) + 35 = -10(0.36) + 7.2 + 35 = -3.6 + 7.2 + 35 = 38.6
$$

So, the vertex is at $$ (0.6, 38.6) $$.

Next, to find the points where the function intersects the x-axis, I need to solve for $$ x $$ when $$ y = 0 $$:

$$
-10x^2 + 12x + 35 = 0
$$

This is a quadratic equation, and I can use the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ to find the roots. Here, $$ a = -10 $$, $$ b = 12 $$, and $$ c = 35 $$.

First, calculate the discriminant:

$$
D = b^2 - 4ac = 12^2 - 4(-10)(35) = 144 + 1400 = 1544
$$

Since the discriminant is positive, there are two real roots:

$$
x = \frac{-12 \pm \sqrt{1544}}{2(-10)} = \frac{-12 \pm 39.29}{-20}
$$

Calculating both possibilities:

1. $$ x = \frac{-12 + 39.29}{-20} = \frac{27.29}{-20} = -1.3645 $$
2. $$ x = \frac{-12 - 39.29}{-20} = \frac{-51.29}{-20} = 2.5645 $$

So, the points of intersection with the x-axis are approximately $$ (-1.36, 0) $$ and $$ (2.56, 0) $$.

Now, to graph the function $$ y = -10x^2 + 12x + 35 $$, I'll plot the vertex at $$ (0.6, 38.6) $$ and the x-intercepts at $$ (-1.36, 0) $$ and $$ (2.56, 0) $$. Since the coefficient of $$ x^2 $$ is negative, the parabola opens downward. I'll also plot a few additional points to get a better shape of the graph:

- For $$ x = -2 $$: $$ y = -10(-2)^2 + 12(-2) + 35 = -40 - 24 + 35 = -29 $$
- For $$ x = 0 $$: $$ y = -10(0)^2 + 12(0) + 35 = 35 $$
- For $$ x = 1 $$: $$ y = -10(1)^2 + 12(1) + 35 = -10 + 12 + 35 = 37 $$
- For $$ x = 3 $$: $$ y = -10(3)^2 + 12(3) + 35 = -90 + 36 + 35 = -19 $$

Plotting these points and connecting them with a smooth curve will give the graph of the quadratic function.

El vértice de la función $$ y = -10x^2 + 12x + 35 $$ se encuentra en el punto $$ (0.6, 38.6) $$. Los puntos de corte con el eje $$ x $$ son aproximadamente $$ (-1.36, 0) $$ y $$ (2.56, 0) $$. La gráfica de la función es una parábola que se abre hacia abajo con su vértice en $$ (0.6, 38.6) $$ y cruza el eje $$ x $$ en los puntos $$ (-1.36, 0) $$ y $$ (2.56, 0) $$.

Hallar las coordenadas del vértice,los puntos de corte con el eje \( x \) y graficar las siguiente función \( y=-10 x^{2}+12 x+35 \) \[ 5: 20 \text { a.m } \]

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