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2 years ago

T(x)=−x^+4x−3 Find two points on the graph of the parabola other than the vertex and x-intercepts.

1 answer:

6 0

The points on the graph of the parabola other than the vertex and x-intercepts where the equation of the parabola is given as t(x) = -x^2 + 4x - 3 are (10, -63) and (5, -8)

<h3>How to determine the points on the graph of the parabola other than the vertex and x-intercepts?</h3>

The equation of the parabola is given as:

t(x) = -x^2 + 4x - 3

The vertex of the parabola is the point where the graph is at the maximum or the minimum

While the x-intercept is the point where the graph crosses the x-axis i.e when y = 0

Having said that, we have the equation of the parabola to be

t(x) = -x^2 + 4x - 3

Set x = 5.

So, we have:

t(5) = -5^2 + 4 * 5 - 3

Evaluate the exponents

t(5) = -25 + 4 * 5 - 3

Evaluate the products

t(5) = -25 + 20 - 3

Evaluate the sum and the difference

t(5) = -8

Set x = 10.

So, we have:

t(10) = -10^2 + 4 * 10 - 3

Evaluate the exponents

t(10) = -100 + 4 * 10 - 3

Evaluate the products

t(10) = -100 + 40 - 3

Evaluate the sum and the difference

t(10) = -63

Hence, the points on the graph of the parabola other than the vertex and x-intercepts where the equation of the parabola is given as t(x) = -x^2 + 4x - 3 are (10, -63) and (5, -8)

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