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3 years ago

Explain how the graph of a quadratic function relates to the solutions of the related quadratic equation.

Mathematics

1 answer:

OleMash [197]3 years ago

8 0

Answer:

The zeros of a quadratic function are the solutions of the related quadratic equation. If the graph of a quadratic function crosses the x-axis, there will be two real number solutions. If the graph of a quadratic function just touches the x-axis, there will be one unique real number, or double root, solution. If the graph of a quadratic function does not cross the x-axis, there will be no real number solution.

Step-by-step explanation:

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Choose the equation below that represents the line passing through the point (1, -4) with a slope of 1/2

tatiyna

For this case we have that by definition, the line equation of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

According to the statement we have:

$m = \frac {1} {2}$

Thus, the equation is of the form:

$y = \frac {1} {2} x + b$

We substitute the given point and find "b":

$-4 = \frac {1} {2} (1) + b\\-4 = \frac {1} {2} + b\\-4- \frac {1} {2} = b\\b = - \frac {9} {2}$

Finally, the equation is of the form:

$y = \frac {1} {2} x- \frac {9} {2}$

Answer:

$y = \frac {1} {2} x- \frac {9} {2}$

6 0

3 years ago

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$