Question

Graph the equation y=x2+4x-5 on the accompanying set of axes. You must plot 5 points including the roots and the vertex.

Answer 1

Roots = (1, 0) and (-5, 0)

Vertex = (-2, -9)

Two additional points = (-4, -5) and (0, -5)

Explanation:

The given equation y = x² + 4x - 5 is a quadratic equation with a positive leading coefficient. Therefore, the shape of the curve is a parabola that opens upwards.

The roots of a quadratic equation are the x-values that satisfy the equation y = 0. They are the points at which the curve crosses the x-axis.

To calculate the roots of the given equation, set it to zero and solve for x.

`\begin{aligned}x^2+4x-5&=0\\x^2+5x-x-5&=0\\x(x+5)-1(x+5)&=0\\(x-1)(x+5)&=0\\\\\implies x-1&=0 \implies x=1\\\implies x+5&=0 \implies x=-5\end{aligned}`

Therefore, the roots of the equation are (1, 0) and (-5, 0).

The x-value of the vertex of a quadratic function in the form ax² + bx + c is -b/2a.

For the given equation y = x² + 4x - 5, a = 1 and b = 4.

To find the x-value of the vertex, substitute these values into the formula:

`x_(\sf vertex)=(-b)/(2a)=(-4)/(2(1))=-2`

To find the y-value of the vertex, substitute the x-value of the vertex into the given equation:

`y_(\sf vertex)=(-2)^2+4(-2)-5=-9`

Therefore, the vertex is (-2, -9).

To find two other points on the curve, substitute two values of x into the equation.

`\begin{aligned}x=-4 \implies y&=(-4)^2+4(-4)-5\\&=16-16-5\\&=0-5\\&=-5\end{aligned}`

`\begin{aligned}x=0 \implies y&=(0)^2+4(0)-5\\&=0+0-5\\&=0-5\\&=-5\end{aligned}`

The axis of symmetry of a quadratic equation is the vertical line that divides the parabola into two mirror images. The x-value of the vertex is the axis of symmetry. Therefore, the axis of symmetry is x = -2.

To graph the given equation:

• Plot the roots (1, 0) and (-5, 0).
• Plot the vertex (-2, -9).
• Plot two additional points (-4, -5) and (0, -5).
• Draw the axis of symmetry x = -2.
• Draw a parabola that is symmetrical about the axis of symmetry and passes through the plotted points.