Question

Find the zeroes of the quadratic polynomial p(x)=x2+x-20 using graph

Answer 1

Answer:

The zeroes are -5 and 4

Step-by-step explanation:

Check the attachment for explanation:-

:)

Answer 2

Concept:

This question requires us to have the knowledge of how to solve a quadratic equation and how to plot a graph of a quadratic equation.

A Graph of a quadratic equation is a parabola.

The number of solution (maximum 2 solution for a quadratic equation) is thhe number of time the curve will meet the axis.

For the quadratic equation y=$a{x}^{2}+bx+c$, if:

if a>0, parabola is upward

if a<0, parabola is downward

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards called the vertex of the parabola.

The vertex is on the axis of symmetry, so its x-coordinate is $-\frac{b}{2a}$.

Given:

The quadratic equation

$p(x)=x^{2}+x-20$

To find:

We need to find the zero of the quadratic equation using graph.

Solution:

We can follow the following steps to plot a graph for the given quadratic equation.

Step 1:

First we need to figure out if the parabola is upward or downward

Since , a>0 therefore parabola is upward

Step 2:

Find the Axis of Symmetry and Vertex of a Parabola.

The equation of the axis of symmetry of the graph of $a{x}^{2}+bx+c$ is$x=-\frac{b}{2a}$

Equation of the axis of symmetry is $-\frac{1}{2}$

Step 3:

Find the x-intercepts and y-intercept.

$y={x}^{2}+x-20$

Substituting $x=-\frac{1}{2}$

$y=(-\frac{1}{2})^{2}+(-\frac{1}{2})-20\\y=\frac{1}{4}-\frac{1}{2}-20\\y=\frac{1-2-80}{4} \\y=-\frac{77}{4}$

x-intercept is$-\frac{1}{2}$

The intercepts are

$(-\frac{1}{2}, -\frac{77}{4} )$

Step 4:

Plot the graph using the above steps

Step 5:

Using this graph find the points where the parabola cuts the x-axis.

We can understand that parabola cuts the graph at (-5,0) and (4,0).

Therefore the zeroes of the graph is -5 and 4.