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elena-s [515]
3 years ago
15

Find the x intercepts of the parabola with vertex (-1,-16) and y intercept (0,-15)

Mathematics
1 answer:
Arlecino [84]3 years ago
8 0
<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>

<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>

<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>

<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>

<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>

<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>

<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>

<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>

<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>

<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>

<span>Or, if you prefer, (3, 0), (-5, 0). </span>
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Write an equation that goes through (8,1) and is perpendicular to 2y + 4x =12
Serggg [28]

Answer:

2y -x + 6 = 0

Step-by-step explanation:

Here a equation and a point is given to us and we are interested in finding a equation which is perpendicular to the given equation and passes through the given point .

The given equation is ,

\sf \longrightarrow 2y + 4x = 12 \\

\sf \longrightarrow 4x + 2y = 12

Firstly convert this into slope intercept form , to find out the slope of the line.

\sf \longrightarrow 2y = -4x +12\\

\sf \longrightarrow y =\dfrac{-4x+12}{2}\\

\sf \longrightarrow y =\dfrac{-4x}{2}+\dfrac{12}{2}\\

\sf \longrightarrow \red{ y = -2x + 6 }

Now on comparing it to slope intercept form which is y = mx + c , we have ,

  • m = -2

And as we know that the product of slopes of two perpendicular lines is -1 . So the slope of the perpendicular line will be negative reciprocal of the slope of the given line. Therefore ,

\sf \longrightarrow m_{\perp}=\dfrac{-1}{-2}=\red{\dfrac{1}{2}}

Now we may use the point slope form of the line to find out the equation of the line using the given point . The point slope form is,

\sf \longrightarrow y - y_1 = m(x - x_1)

Now on substituting the respective values we have,

\sf \longrightarrow y - 1 = \dfrac{1}{2}(x-8)\\

\sf \longrightarrow 2(y-1 )= x -8 \\

\sf \longrightarrow 2y -2=x-8\\

\sf \longrightarrow \underline{\boxed{\bf 2y - x + 6=0}}

5 0
3 years ago
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