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2 years ago

Identify the vertex of the quadratic function in standard form y=2x^2 -16x+31

Mathematics

1 answer:

lapo4ka [179]2 years ago

8 0

Answer:

(4,1)

Step-by-step explanation:

The vertex of the quadratic can be found using -b/2a. The vertex is the point (-b/2a, f(-b/2a)).

To find the x coordinate of the quadratic, define its a,b, and c. The quadratic has a = 2, b=-16, and c=31. Then -b/2a is -(-16)/2(2) = 16/4 = 4.

To find the y-coordinate, input the x=4 and solve for y.

$y=2(4)^2-16(4)+31\\y=2(16)-64+31\\y=32-64+31\\y=1$

The vertex is (4,1).

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Question:

Find the number of real number solutions for the equation. x^2 + 5x + 7 = 0

Answer:

The number of real solutions for the equation $x^{2}+5 x+7=0$ is zero

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For a Quadratic Equation of form : $a x^{2}+b x+c=0$ ---- eqn 1

The solution is $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now , the given Quadratic Equation is $x^{2}+5 x+7=0$ ---- eqn 2

On comparing Equation (1) and Equation(2), we get

a = 1 , b = 5 and c = 7

In $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ , $b^2 - 4ac$ is called the discriminant of the quadratic equation

Its value determines the nature of roots

Now, here are the rules with discriminants:

1) D > 0; there are 2 real solutions in the equation

2) D = 0; there is 1 real solution in the equation

3) D < 0; there are no real solutions in the equation

Now let solve for given equation

$D= b^2 - 4ac\\\\D = 5^2 - 4(1)(7)\\\\D = 25 - 28 \\\\D = -3$

Since -3 is less than 0, this means that there are 0 real solutions in this equation.

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