2y+5=1? Solve for A.

Mathematics

2 answers:

shepuryov [24]3 years ago

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aalyn [17]3 years ago

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Answer:

2y + 5 = 1

⇛2y = 1 - 5

⇛2y = -4

⇛y = -4 ÷ 2

⇛y = -4/2

⇛y = -2

Therefore, the value of "y" is -2.

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Part A

$4x^2 -7x-15=0\\\\(4x+5)(x-3)=0\\\\x=-\frac{5}{4}, 3$

So, the x-intercepts are $\boxed{\left(-\frac{5}{4}, 0 \right), (3,0)}$

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The vertex will be a minimum because the coefficient of $x^2$ is positive.

The x-coordinate of the vertex is $x=-\frac{-7}{2(4)}=\frac{7}{8}$

Substituting this back into the function, we get $f\left(\frac{7}{8} \right)=4\left(\frac{7}{8} \right)^2 -7\left(\frac{7}{8} \right)^2 -15=-\frac{289}{16}$