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Pavel [41]
3 years ago
8

Find an equation of the parabola y = ax2 + bx + c that passes through (0, 1) and is tangent to the line y = 4x − 4 at (1, 0).

Mathematics
1 answer:
larisa86 [58]3 years ago
6 0
We are given the following:
          - parabola passes to both (1,0) and (0,1) 
<span>          - slope at x = 1 is 4 from the equation of the tangent line </span>

<span>First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. W</span><span>hen x = 0, y = 1. So, c should be equal to 1. The</span><span> parabola is y = ax^2 + bx + 1 </span>

<span>Now, we can substitute the point (1,0) into the equation,
</span>0 = a(1)^2 + b(1) + 1 
<span>0 = a + b + 1
a + b = -1 </span>

<span>The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.</span>
<span>We take the derivative of the  equation ,
y = ax^2 + bx + 1</span>
<span>y' = 2ax + b 
</span>
<span>x = 1, y' = 2
</span>4 = 2a(1) + b 
<span>4 = 2a + b </span>

So, we have two equations and  two unknowns,<span> </span>
<span>2a + b = 4 </span>
<span>a + b = -1 
</span><span>
Solving simultaneously,
a = 5 </span>
<span>b = -6</span>

<span>Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .</span>
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