Maths-

General

Easy

Question

# Estimate the coordinates of the vertex of the graph of f(x) = 1.25x^{2} -2x -1 below. Then explain how to find the exact coordinates

Hint:

### For a quadratic function is in standard form, f(x)=ax2+bx+c.

A vertical line passing through the vertex is called the axis of symmetry for the parabola.

Axis of symmetry x=−b/2a

Vertex The vertex of the parabola is located at a pair of coordinates which we will call (*h, k*). where h is value of x in axis of symmetry formula and k is f(h).

## The correct answer is: (0.8 , -1.8)

### We have given a function

f(x) = 1.25x^{2} -2x -1

This quadratic function is in standard form, f(x)=ax^{2}+bx+c.

For every quadratic function in standard form the axis of symmetry is given by the formula x=−b/2a.

In f(x)= 1.25x^{2} -2x -1, a= 1.25, b= -2, and c= -1. So, the equation for the axis of symmetry is given by

x = −(-2)/2(1.25)

x = 2/2.5

x = 0.8

The equation of the axis of symmetry for f(x)= 1.25x^{2} -2x -1 is x = 0.8.

The x coordinate of the vertex is the same:

h = 0.8

The y coordinate of the vertex is :

k = f(h)

k = 1.25h^{2} -2h -1

k = 1.25(0.8)^{2} - 2(0.8) - 1

k = 0.8 - 1.6 – 1

k = -1.8

Therefore, the vertex is (0.8 , -1.8)

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