Question

graph the parabola.

$y = -4x^2 + 5$

plot five points on the parabola: the vertex, two points to the left of the vertex, and two points to the right of the vertex. then click on the graph-a-function button.

Explanation:

Step1: Identify the vertex form

The given parabola equation is \( y = -4x^2 + 5 \). For a parabola in the form \( y = ax^2 + bx + c \), the x - coordinate of the vertex is \( x = -\frac{b}{2a} \). Here, \( a=-4 \) and \( b = 0 \), so \( x=-\frac{0}{2\times(-4)} = 0 \). Substitute \( x = 0 \) into the equation: \( y=-4(0)^2 + 5=5 \). So the vertex is \( (0,5) \).

Step2: Find points to the left of the vertex (x < 0)

Let's take \( x=-1 \): Substitute into \( y=-4x^2 + 5 \), \( y=-4(-1)^2+5=-4 + 5 = 1 \). So the point is \( (-1,1) \).
Let's take \( x = - 2 \): Substitute into \( y=-4x^2 + 5 \), \( y=-4(-2)^2+5=-4\times4 + 5=-16 + 5=-11 \). So the point is \( (-2,-11) \).

Step3: Find points to the right of the vertex (x>0)

Let's take \( x = 1 \): Substitute into \( y=-4x^2 + 5 \), \( y=-4(1)^2+5=-4 + 5 = 1 \). So the point is \( (1,1) \).
Let's take \( x = 2 \): Substitute into \( y=-4x^2 + 5 \), \( y=-4(2)^2+5=-16 + 5=-11 \). So the point is \( (2,-11) \).

Now we can plot the vertex \( (0,5) \), the left - side points \( (-1,1) \), \( (-2,-11) \) and the right - side points \( (1,1) \), \( (2,-11) \) and then draw the parabola.

Answer:

The vertex is \((0,5)\), left - side points: \((-1,1)\), \((-2,-11)\); right - side points: \((1,1)\), \((2,-11)\) (these points can be used to graph the parabola \(y=-4x^2 + 5\))